gr-gifted-amateur4-2-b68dff55-6015-425e-ab90-99e56b904fff

30.3 Two-dimensional surfaces

So far we have considered the curvature of lines in three dimensions. We now upgrade the discussion to deal with the curvature of twodimensional surfaces embedded in flat three-dimensional space. For onedimensional surfaces (i.e. curves) we started by comparing the behaviour near closely spaced points to that of points on a circle (and then, subsequently, on a parabola). For the two-dimensional surface, we might imagine a sphere would be the measure of curvature. However, we can't say that the curvature will be the same in two different directions, since this is generally not true. In fact, we must evaluate two curvature constants, κ 1 κ 1 kappa_(1)\kappa_{1}κ1 and κ 2 κ 2 kappa_(2)\kappa_{2}κ2, to characterize the curvature of a plane. It seems like a tough problem to determine these two constants, since we don't know in which directions to evaluate the curvature. The task is dramatically simplified by a discovery by Leonhard Euler. 10 10 ^(10){ }^{10}10 If the two curvature constants aren't equal then there will be some direction where the curvature with be a minimum and another where it is a maximum. Euler's discovery was that these directions are perpendicular. As a result, just as the curvature of a line is measured by the radius of a circle, the curvature of the two-dimensional plane is measured using the major and minor axes of an ellipse.
We start with the coordinate approach, where we measure curvature by comparing the height of a curved surface above a plane which forms the tangent to the surface at some nearby point (Fig. 30.6). Taking our lead from our analysis of the curve, in two dimensions we have that the height z z zzz of the curved surface, written in terms of coordinates x i x i x^(i)x^{i}xi in the tangent plane, is
(30.17) z = 1 2 K i j x i x j , (30.17) z = 1 2 K i j x i x j , {:(30.17)z=(1)/(2)K_(ij)x^(i)x^(j)",":}\begin{equation*} z=\frac{1}{2} K_{i j} x^{i} x^{j}, \tag{30.17} \end{equation*}(30.17)z=12Kijxixj,
where K i j K i j K_(ij)K_{i j}Kij are the components of a 2 × 2 2 × 2 2xx22 \times 22×2 real symmetric tensor. This tensor is the key to understanding the classical curvature of curved surfaces since it defines the ellipse whose major and minor axes are the eigenvalues of K i j K i j K_(ij)K_{i j}Kij. The inverse of these eigenvalues will give the two components of curvature, as we shall see below.
Example 30.5
Consider a surface embedded in three-dimensional space. The departure z z zzz of the curved surface from the flat one can be expanded as
(30.18) z = 1 2 a x 2 + b x y + 1 2 c y 2 (30.18) z = 1 2 a x 2 + b x y + 1 2 c y 2 {:(30.18)z=(1)/(2)*ax^(2)+bxy+(1)/(2)cy^(2):}\begin{equation*} z=\frac{1}{2} \cdot a x^{2}+b x y+\frac{1}{2} c y^{2} \tag{30.18} \end{equation*}(30.18)z=12ax2+bxy+12cy2
or, equivalently,
(30.19) z = 1 2 ( x y ) ( a b b c ) ( x y ) . (30.19) z = 1 2 x y a b b c ( x y ) . {:(30.19)z=(1)/(2)([x,y])([a,b],[b,c])((x)/(y)).:}z=\frac{1}{2}\left(\begin{array}{ll} x & y \end{array}\right)\left(\begin{array}{ll} a & b \tag{30.19}\\ b & c \end{array}\right)\binom{x}{y} .(30.19)z=12(xy)(abbc)(xy).
It is always possible make a coordinate transformation, effectively by rotating the axes
(30.20) ( x y ) = ( cos α sin α sin α cos α ) ( ξ η ) . (30.20) ( x y ) = cos α sin α sin α cos α ( ξ η ) . {:(30.20)((x)/(y))=([cos alpha,sin alpha],[-sin alpha,cos alpha])((xi )/(eta)).:}\binom{x}{y}=\left(\begin{array}{cc} \cos \alpha & \sin \alpha \tag{30.20}\\ -\sin \alpha & \cos \alpha \end{array}\right)\binom{\xi}{\eta} .(30.20)(xy)=(cosαsinαsinαcosα)(ξη).
We choose the angle α α alpha\alphaα to diagonalize the matrix, such that we have
(30.21) z = 1 2 ( ξ η ) ( κ 1 0 0 κ 2 ) ( ξ η ) (30.21) z = 1 2 ξ η κ 1 0 0 κ 2 ( ξ η ) {:(30.21)z=(1)/(2)([xi,eta])([kappa_(1),0],[0,kappa_(2)])((xi )/(eta)):}z=\frac{1}{2}\left(\begin{array}{ll} \xi & \eta \end{array}\right)\left(\begin{array}{cc} \kappa_{1} & 0 \tag{30.21}\\ 0 & \kappa_{2} \end{array}\right)\binom{\xi}{\eta}(30.21)z=12(ξη)(κ100κ2)(ξη)
or equivalently
(30.22) z = 1 2 κ 1 ξ 2 + 1 2 κ 2 η 2 (30.22) z = 1 2 κ 1 ξ 2 + 1 2 κ 2 η 2 {:(30.22)z=(1)/(2)kappa_(1)xi^(2)+(1)/(2)kappa_(2)eta^(2):}\begin{equation*} z=\frac{1}{2} \kappa_{1} \xi^{2}+\frac{1}{2} \kappa_{2} \eta^{2} \tag{30.22} \end{equation*}(30.22)z=12κ1ξ2+12κ2η2
The new variables, ξ ξ xi\xiξ and η η eta\etaη are the coordinates on the principal axes of the ellipse. In this context, the eigenvalues κ i κ i kappa_(i)\kappa_{i}κi are called the principal curvatures and can be written in terms of radii of curvature ρ i ρ i rho_(i)\rho_{i}ρi as κ 1 = 1 / ρ 1 κ 1 = 1 / ρ 1 kappa_(1)=1//rho_(1)\kappa_{1}=1 / \rho_{1}κ1=1/ρ1 and κ 2 = 1 / ρ 2 κ 2 = 1 / ρ 2 kappa_(2)=1//rho_(2)\kappa_{2}=1 / \rho_{2}κ2=1/ρ2. The principal axes give us the largest (major) and smallest (minor) widths of the ellipse, which provide our measure of curvature.
Let's apply this approach to a spherical surface of radius a a aaa. For the surface balanced on a plane we can write z = ( a 2 + x 2 + y 2 ) 1 2 + a z = a 2 + x 2 + y 2 1 2 + a z=-(a^(2)+x^(2)+y^(2))^((1)/(2))+az=-\left(a^{2}+x^{2}+y^{2}\right)^{\frac{1}{2}}+az=(a2+x2+y2)12+a. Near the point of contact, we have z 1 2 a 2 ( x 2 + y 2 ) z 1 2 a 2 x 2 + y 2 z~~(1)/(2a^(2))*(x^(2)+y^(2))z \approx \frac{1}{2 a^{2}} \cdot\left(x^{2}+y^{2}\right)z12a2(x2+y2) and so we have two principal curvatures of 1 / a 1 / a 1//a1 / a1/a, both with radius of curvature a a aaa, just as we would expect.
11 11 ^(11){ }^{11}11 Gauss' own version can be found in his 1827 paper Disquisitiones generales circa superficies curvas (General Investigations of Curved Surfaces). A translation of the paper along with a running lation of the paper along with a running
commentary given in terms of modern commentary given in terms of modern
mathematical notation can be found mathematical notation can be found
in Volume II of Michael Spivak's A A AAA Comprehensive Introduction to Differential Geometry and comes highly recommended.
12 12 ^(12){ }^{12}12 Modern English usage of the word egregious, which formerly meant 'remarkable, usually implies that something is remarkably bad. H. W. Fowler (1858-1933) notes that it is especially (1858-1933) notes that it is especially
applied to the nouns "ass, coxcomb, applied to the nouns "ass, coxcomb,
liar, imposter, folly, blunder, waste", liar, imposter, folly, blunder, waste",
and that "Reversion to the original sense ... is mere pedantry."
13 13 ^(13){ }^{13}13 See the excellent book by Needham (2021) for more details and several proofs. Needham makes sense of the theorem which, in Gauss' original form (discussed here), admittedly looks like it was plucked magically from the air.
Here's a look at Gauss' remarkable result. 13 13 ^(13){ }^{13}13 In two dimensions, the flat, Euclidean plane is described by a line element
(30.23) d s 2 = ( d ξ 1 ) 2 + ( d ξ 2 ) 2 (30.23) d s 2 = d ξ 1 2 + d ξ 2 2 {:(30.23)ds^(2)=(dxi^(1))^(2)+(dxi^(2))^(2):}\begin{equation*} \mathrm{d} s^{2}=\left(\mathrm{d} \xi^{1}\right)^{2}+\left(\mathrm{d} \xi^{2}\right)^{2} \tag{30.23} \end{equation*}(30.23)ds2=(dξ1)2+(dξ2)2
Gauss assumed that in a sufficiently small (that is, infinitesimal) patch of a curved space, it would always be possible to describe the space using such a coordinate system. However, a general two-dimensional curved space will not be able to be described by the (flat-space) ξ i ξ i xi^(i)\xi^{i}ξi coordinates over a finite neighbourhood. Instead, there will be a perfectly good coordinate system ( x 1 , x 2 ) x 1 , x 2 (x^(1),x^(2))\left(x^{1}, x^{2}\right)(x1,x2) that does cover the curved space. In this latter coordinate system, the line element is written as (just as we've had previously)
(30.24) d s 2 = g i j d x i d x j (30.24) d s 2 = g i j d x i d x j {:(30.24)ds^(2)=g_(ij)dx^(i)dx^(j):}\begin{equation*} \mathrm{d} s^{2}=g_{i j} \mathrm{~d} x^{i} \mathrm{~d} x^{j} \tag{30.24} \end{equation*}(30.24)ds2=gij dxi dxj
As usual, the values of g i j g i j g_(ij)g_{i j}gij depend on the particular coordinate system chosen (although it will turn out that they also encode the intrinsic properties of the space).
14 14 ^(14){ }^{14}14 This makes it similar to s s sss in the one-
dimensional case.
Gauss sought to define a measure of curvature in terms of a function of the components of the metric of g i j g i j g_(ij)g_{i j}gij and their derivatives that depends only on the intrinsic properties of the space and not on the particular coordinate system chosen. The properties of the space and not on the particular coordinate system chosen. The
formidable equation that Gauss discovered is (employing the comma notation for derivatives)
K ( x 1 , x 2 ) = 1 2 g [ 2 g 12 , 12 g 11 , 22 g 22 , 11 ] g 22 4 g 2 [ g 11 , 1 ( 2 g 12 , 2 g 22 , 1 ) g 11 , 2 2 ] + g 12 4 g 2 [ g 11 , 1 g 22 , 2 2 g 11 , 2 g 22 , 1 + ( 2 g 12 , 1 g 11 , 2 ) ( 2 g 12 , 2 g 22 , 1 ) ] (30.25) g 11 4 g 2 [ g 22 , 2 ( 2 g 12 , 1 g 11 , 2 ) g 22 , 1 2 ] K x 1 , x 2 = 1 2 g 2 g 12 , 12 g 11 , 22 g 22 , 11 g 22 4 g 2 g 11 , 1 2 g 12 , 2 g 22 , 1 g 11 , 2 2 + g 12 4 g 2 g 11 , 1 g 22 , 2 2 g 11 , 2 g 22 , 1 + 2 g 12 , 1 g 11 , 2 2 g 12 , 2 g 22 , 1 (30.25) g 11 4 g 2 g 22 , 2 2 g 12 , 1 g 11 , 2 g 22 , 1 2 {:[K(x^(1),x^(2))=(1)/(2g)*[2g_(12,12)-g_(11,22)-g_(22,11)]],[-(g_(22))/(4g^(2))*[g_(11,1)(2g_(12,2)-g_(22,1))-g_(11,2)^(2)]],[+(g_(12))/(4g^(2))*[g_(11,1)g_(22,2)-2g_(11,2)g_(22,1)+(2g_(12,1)-g_(11,2))(2g_(12,2)g_(22,1))]],[(30.25)-(g_(11))/(4g^(2))[g_(22,2)(2g_(12,1)-g_(11,2))-g_(22,1)^(2)]]:}\begin{align*} K\left(x^{1}, x^{2}\right)= & \frac{1}{2 g} \cdot\left[2 g_{12,12}-g_{11,22}-g_{22,11}\right] \\ & -\frac{g_{22}}{4 g^{2}} \cdot\left[g_{11,1}\left(2 g_{12,2}-g_{22,1}\right)-g_{11,2}^{2}\right] \\ & +\frac{g_{12}}{4 g^{2}} \cdot\left[g_{11,1} g_{22,2}-2 g_{11,2} g_{22,1}+\left(2 g_{12,1}-g_{11,2}\right)\left(2 g_{12,2} g_{22,1}\right)\right] \\ & -\frac{g_{11}}{4 g^{2}}\left[g_{22,2}\left(2 g_{12,1}-g_{11,2}\right)-g_{22,1}^{2}\right] \tag{30.25} \end{align*}K(x1,x2)=12g[2g12,12g11,22g22,11]g224g2[g11,1(2g12,2g22,1)g11,22]+g124g2[g11,1g22,22g11,2g22,1+(2g12,1g11,2)(2g12,2g22,1)](30.25)g114g2[g22,2(2g12,1g11,2)g22,12]
where g g ggg is the determinant
(30.26) g ( x 1 , x 2 ) = g 11 g 22 g 12 2 (30.26) g x 1 , x 2 = g 11 g 22 g 12 2 {:(30.26)g(x_(1),x_(2))=g_(11)g_(22)-g_(12)^(2):}\begin{equation*} g\left(x_{1}, x_{2}\right)=g_{11} g_{22}-g_{12}^{2} \tag{30.26} \end{equation*}(30.26)g(x1,x2)=g11g22g122
The significance of this is that someone in possession of the components of the metric also has a failsafe means of calculating the curvature of the surface that cannot be the consequence merely of a particular choice of coordinates. Gauss was, understandably, elated to have discovered this wonderful result.
As a concrete example, we have for the sphere that g θ θ = a 2 g θ θ = a 2 g_(theta theta)=a^(2)g_{\theta \theta}=a^{2}gθθ=a2 and g ϕ ϕ = a 2 sin 2 θ g ϕ ϕ = a 2 sin 2 θ g_(phi phi)=a^(2)sin^(2)thetag_{\phi \phi}=a^{2} \sin ^{2} \thetagϕϕ=a2sin2θ. We find that
(30.27) K ( θ , ϕ ) = 1 a 2 , (30.27) K ( θ , ϕ ) = 1 a 2 , {:(30.27)K(theta","phi)=(1)/(a^(2))",":}\begin{equation*} K(\theta, \phi)=\frac{1}{a^{2}}, \tag{30.27} \end{equation*}(30.27)K(θ,ϕ)=1a2,
which is to say that the curvature function is 1 / ( radius ) 2 1 / (  radius  ) 2 1//(" radius ")^(2)1 /(\text { radius })^{2}1/( radius )2, which is only given in terms of a coordinate of the space itself, its radius. It is also heartening that the flat, Euclidean plane gives
(30.28) K ( ξ 1 , ξ 2 ) = 0 (30.28) K ξ 1 , ξ 2 = 0 {:(30.28)K(xi^(1),xi^(2))=0:}\begin{equation*} K\left(\xi^{1}, \xi^{2}\right)=0 \tag{30.28} \end{equation*}(30.28)K(ξ1,ξ2)=0
We will have more to say about Gaussian curvature but, happily won't have to make use of Gauss' remarkable equation as there are several alternative ways of extracting K K KKK.

30.4 Gauss' equation

In one dimension, we considered lengths of lines, so it would seem natural to look at areas in two dimensions. In this spirit, we can define the curvature of a surface at a point P P P\mathcal{P}P as
(30.29) K ( P ) = d Ω d A = lim Δ A 0 ( dimensionless area swept out on unit sphere by normals ) ( corresponding area Δ A on actual surface ) (30.29) K ( P ) = d Ω d A = lim Δ A 0 (  dimensionless area swept out   on unit sphere by normals  ) (  corresponding area  Δ A  on actual surface  ) {:(30.29)K(P)=(dOmega)/((d)A)=lim_(Delta A rarr0)(((" dimensionless area swept out ")/(" on unit sphere by normals ")))/(((" corresponding area "Delta A)/(" on actual surface "))):}\begin{equation*} K(\mathcal{P})=\frac{\mathrm{d} \Omega}{\mathrm{~d} A}=\lim _{\Delta A \rightarrow 0} \frac{\binom{\text { dimensionless area swept out }}{\text { on unit sphere by normals }}}{\binom{\text { corresponding area } \Delta A}{\text { on actual surface }}} \tag{30.29} \end{equation*}(30.29)K(P)=dΩ dA=limΔA0( dimensionless area swept out  on unit sphere by normals )( corresponding area ΔA on actual surface )
Turning to the description in terms of vectors, we define a twodimensional surface embedded in flat three-dimensional space by a vector-valued function X ( x 1 , x 2 ) X x 1 , x 2 X(x^(1),x^(2))\boldsymbol{X}\left(x^{1}, x^{2}\right)X(x1,x2), where the coordinates x 1 x 1 x_(1)x_{1}x1 and x 2 x 2 x_(2)x_{2}x2 parametrize the surface [i.e. the coordinate system ( x 1 , x 2 ) x 1 , x 2 (x^(1),x^(2))\left(x^{1}, x^{2}\right)(x1,x2) lies within the surface 14 ] 14 {:^(14)]\left.{ }^{14}\right]14]. That is, given a point with coordinates in the surface of x 1 x 1 x^(1)x^{1}x1 and x 2 x 2 x^(2)x^{2}x2, the vector-valued function X X X\boldsymbol{X}X returns the vector from the origin to the surface at that point. The two-dimensional surface has two basis vectors, which are given by
(30.30) e 1 = X x 1 , e 2 = X x 2 (30.30) e 1 = X x 1 , e 2 = X x 2 {:(30.30)e_(1)=(del X)/(delx^(1))","quade_(2)=(del X)/(delx^(2)):}\begin{equation*} e_{1}=\frac{\partial \boldsymbol{X}}{\partial x^{1}}, \quad e_{2}=\frac{\partial \boldsymbol{X}}{\partial x^{2}} \tag{30.30} \end{equation*}(30.30)e1=Xx1,e2=Xx2
Example 30.7
Let's now use this expression to derive the Gaussian curvature using these two basis vectors. The normal to the surface is n = e 1 × e 2 n = e 1 × e 2 n=e_(1)xxe_(2)n=e_{1} \times e_{2}n=e1×e2 and so the unit normal 15 15 ^(15){ }^{15}15 is
(30.31) n ^ = ( e 1 × e 2 ) | e 1 × e 2 | (30.31) n ^ = e 1 × e 2 e 1 × e 2 {:(30.31) hat(n)=((e_(1)xxe_(2)))/(|e_(1)xxe_(2)|):}\begin{equation*} \hat{n}=\frac{\left(e_{1} \times e_{2}\right)}{\left|e_{1} \times e_{2}\right|} \tag{30.31} \end{equation*}(30.31)n^=(e1×e2)|e1×e2|
An element of area on the surface is given by
(30.32) d A = | e 1 × e 2 | d x 1 d x 2 = | X x 1 × X x 2 | x 1 x 2 (30.32) d A = e 1 × e 2 d x 1 d x 2 = X x 1 × X x 2 x 1 x 2 {:(30.32)dA=|e_(1)xxe_(2)|dx^(1)dx^(2)=|(del X)/(delx^(1))xx(del X)/(delx^(2))|delx^(1)delx^(2):}\begin{equation*} \mathrm{d} A=\left|\boldsymbol{e}_{1} \times \boldsymbol{e}_{2}\right| \mathrm{d} x^{1} \mathrm{~d} x^{2}=\left|\frac{\partial \boldsymbol{X}}{\partial x^{1}} \times \frac{\partial \boldsymbol{X}}{\partial x^{2}}\right| \partial x^{1} \partial x^{2} \tag{30.32} \end{equation*}(30.32)dA=|e1×e2|dx1 dx2=|Xx1×Xx2|x1x2
For the area swept out by the unit normals, we note that it will be proportional to
(30.33) d Ω | n ^ x 1 × n ^ x 2 | d x 1 d x 2 (30.33) d Ω n ^ x 1 × n ^ x 2 d x 1 d x 2 {:(30.33)dOmega prop|(del( hat(n)))/(delx^(1))xx(del( hat(n)))/(delx^(2))|dx^(1)dx^(2):}\begin{equation*} \mathrm{d} \Omega \propto\left|\frac{\partial \hat{\boldsymbol{n}}}{\partial x^{1}} \times \frac{\partial \hat{\boldsymbol{n}}}{\partial x^{2}}\right| \mathrm{d} x^{1} \mathrm{~d} x^{2} \tag{30.33} \end{equation*}(30.33)dΩ|n^x1×n^x2|dx1 dx2
This makes it parallel to n ^ n ^ hat(n)\hat{\boldsymbol{n}}n^ itself. To pick out the sign, we dot it with n ^ n ^ hat(n)\hat{\boldsymbol{n}}n^ to obtain
(30.34) d Ω = n ^ ( n ^ x 1 × n ^ x 2 ) d x 1 d x 2 (30.34) d Ω = n ^ n ^ x 1 × n ^ x 2 d x 1 d x 2 {:(30.34)dOmega= hat(n)*((del( hat(n)))/(delx^(1))xx(del( hat(n)))/(delx^(2)))dx^(1)dx^(2):}\begin{equation*} \mathrm{d} \Omega=\hat{\boldsymbol{n}} \cdot\left(\frac{\partial \hat{\boldsymbol{n}}}{\partial x^{1}} \times \frac{\partial \hat{\boldsymbol{n}}}{\partial x^{2}}\right) \mathrm{d} x^{1} \mathrm{~d} x^{2} \tag{30.34} \end{equation*}(30.34)dΩ=n^(n^x1×n^x2)dx1 dx2
The Gaussian curvature is then given by
(30.35) K = d Ω d A = n ^ ( n ^ x I × n ^ ^ x 2 ) | e 1 × e 2 | (30.35) K = d Ω d A = n ^ n ^ x I × n ^ ^ x 2 e 1 × e 2 {:(30.35)K=(dOmega)/((d)A)=(( hat(n))*((del( hat(n)))/(delx^(I))xx(del( hat(hat(n))))/(delx^(2))))/(|e_(1)xxe_(2)|):}\begin{equation*} K=\frac{\mathrm{d} \Omega}{\mathrm{~d} A}=\frac{\hat{\boldsymbol{n}} \cdot\left(\frac{\partial \hat{\boldsymbol{n}}}{\partial x^{\mathrm{I}}} \times \frac{\partial \hat{\hat{n}}}{\partial x^{2}}\right)}{\left|\boldsymbol{e}_{1} \times \boldsymbol{e}_{2}\right|} \tag{30.35} \end{equation*}(30.35)K=dΩ dA=n^(n^xI×n^^x2)|e1×e2|
We can now compute the curvature of a surface from knowledge of its basis vectors.
Although this is all very interesting, an approach even more closely based on modern differential geometry will be more useful to us. Consider the 3 -vector e ν e ν e_(nu)\boldsymbol{e}_{\nu}eν and its derivative e ν x μ e ν x μ (dele_(nu))/(delx^(mu))\frac{\partial \boldsymbol{e}_{\nu}}{\partial x^{\mu}}eνxμ. In general, we can choose to write the components of the derivative in the form
(30.36) e ν x μ = Γ μ ν λ e λ + K μ ν n ^ (30.36) e ν x μ = Γ μ ν λ e λ + K μ ν n ^ {:(30.36)(dele_(nu))/(delx^(mu))=Gamma_(mu nu)^(lambda)e_(lambda)+K_(mu nu) hat(n):}\begin{equation*} \frac{\partial \boldsymbol{e}_{\nu}}{\partial x^{\mu}}=\Gamma_{\mu \nu}^{\lambda} \boldsymbol{e}_{\lambda}+K_{\mu \nu} \hat{\boldsymbol{n}} \tag{30.36} \end{equation*}(30.36)eνxμ=Γμνλeλ+Kμνn^
This has a part expressed in terms on vectors confined to the surface (the first term on the right) and a part expressed in terms of the bit sticking out of the surface (the second term). Since n ^ n ^ = 1 n ^ n ^ = 1 hat(n)* hat(n)=1\hat{\boldsymbol{n}} \cdot \hat{\boldsymbol{n}}=1n^n^=1 and n ^ e ν = 0 n ^ e ν = 0 hat(n)*e_(nu)=0\hat{\boldsymbol{n}} \cdot \boldsymbol{e}_{\nu}=0n^eν=0, we also find an expression for the matrix K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν
(30.37) K μ ν = e ν x μ n ^ (30.37) K μ ν = e ν x μ n ^ {:(30.37)K_(mu nu)=(dele_(nu))/(delx^(mu))* hat(n):}\begin{equation*} K_{\mu \nu}=\frac{\partial \boldsymbol{e}_{\nu}}{\partial x^{\mu}} \cdot \hat{\boldsymbol{n}} \tag{30.37} \end{equation*}(30.37)Kμν=eνxμn^
which is known 16 16 ^(16){ }^{16}16 as Gauss' equation. The matrix K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν is designed to look like the Gaussian curvature function K ( x 1 , x 2 ) K x 1 , x 2 K(x_(1),x_(2))K\left(x_{1}, x_{2}\right)K(x1,x2), but it is not yet clear how. We look into this in the next section.
Example 30.8
Let's compute the components of K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν for a cylinder. The points on a cylinder's surface are described by vectors, with Cartesian components given by
(30.38) X ( θ , z ) = ( a cos θ a sin θ z ) (30.38) X ( θ , z ) = a cos θ a sin θ z {:(30.38)X(theta","z)=([a cos theta],[a sin theta],[z]):}\boldsymbol{X}(\theta, z)=\left(\begin{array}{c} a \cos \theta \tag{30.38}\\ a \sin \theta \\ z \end{array}\right)(30.38)X(θ,z)=(acosθasinθz)
15 15 ^(15){ }^{15}15 It is the unit normal that is the useful quantity in this context, so we divide by the magnitude n = e 1 × e 2 n = e 1 × e 2 n=e_(1)xxe_(2)∣n=e_{1} \times e_{2} \midn=e1×e2. There are several cases below where we differentiate n ^ n ^ hat(n)\hat{\boldsymbol{n}}n^. However, although | n | | n | |n||\boldsymbol{n}||n| can depend on the coordinates, we don't differentiate this normalization factor. Rather we scale n n n\boldsymbol{n}n by a factor 1 / | n | 1 / | n | 1//|n|1 /|\boldsymbol{n}|1/|n| at some position to make it a unit vector, and then treat the factor as a constant.
16 16 ^(16){ }^{16}16 Gauss came up with a lot of equations, and so this name doesn't uniquely identify it!
17 17 ^(17){ }^{17}17 For later use, we also have metric components g μ ν = e μ e ν g μ ν = e μ e ν g_(mu nu)=e_(mu)*e_(nu)g_{\mu \nu}=\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}gμν=eμeν of g θ θ = a 2 g θ θ = a 2 g_(theta theta)=a^(2)g_{\theta \theta}=a^{2}gθθ=a2 and g z z = 1 g z z = 1 g_(zz)=1g_{z z}=1gzz=1
Fig. 30.7 A curve, passing through point P P P\mathcal{P}P, embedded in a twodimensional surface.
18 18 ^(18){ }^{18}18 Some terminology in classical geometry includes the first fundamental form, defined as
(30.45) F 1 = g μ ν d X μ d X ν (30.45) F 1 = g μ ν d X μ d X ν {:(30.45)F_(1)=g_(mu nu)dX^(mu)dX^(nu):}\begin{equation*} F_{1}=g_{\mu \nu} \mathrm{d} X^{\mu} \mathrm{d} X^{\nu} \tag{30.45} \end{equation*}(30.45)F1=gμνdXμdXν
This is simply d s 2 d s 2 ds^(2)\mathrm{d} s^{2}ds2 that we've written before. The second fundamental form is found by considering κ n = κ n = kappa_(n)=\kappa_{n}=κn= K μ ν t μ t ν = K μ ν ( d X μ / d s ) ( d X ν / d s ) K μ ν t μ t ν = K μ ν d X μ / d s d X ν / d s K_(mu nu)t^(mu)t^(nu)=K_(mu nu)(dX^(mu)//ds)(dX^(nu)//ds)K_{\mu \nu} t^{\mu} t^{\nu}=K_{\mu \nu}\left(\mathrm{d} X^{\mu} / \mathrm{d} s\right)\left(\mathrm{d} X^{\nu} / \mathrm{d} s\right)Kμνtμtν=Kμν(dXμ/ds)(dXν/ds), and removing the path elements d s d s ds\mathrm{d} sds to write
F 2 = K μ ν d X μ d X ν . F 2 = K μ ν d X μ d X ν . F_(2)=K_(mu nu)dX^(mu)dX^(nu).F_{2}=K_{\mu \nu} \mathrm{d} X^{\mu} \mathrm{d} X^{\nu} .F2=KμνdXμdXν.
(30.46)
We won't have cause to use this terminology.
The basis vectors and unit normal are found to be 17 17 ^(17){ }^{17}17
(30.39) e θ = ( a sin θ a cos θ 0 ) , e z = ( 0 0 1 ) , n ^ = ( cos θ sin θ 0 ) (30.39) e θ = a sin θ a cos θ 0 , e z = 0 0 1 , n ^ = cos θ sin θ 0 {:(30.39)e_(theta)=([-a sin theta],[a cos theta],[0])","quade_(z)=([0],[0],[1])","quad hat(n)=([cos theta],[sin theta],[0]):}\boldsymbol{e}_{\theta}=\left(\begin{array}{c} -a \sin \theta \tag{30.39}\\ a \cos \theta \\ 0 \end{array}\right), \quad \boldsymbol{e}_{z}=\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right), \quad \hat{\boldsymbol{n}}=\left(\begin{array}{c} \cos \theta \\ \sin \theta \\ 0 \end{array}\right)(30.39)eθ=(asinθacosθ0),ez=(001),n^=(cosθsinθ0)
Differentiating, the only non-zero component results from
(30.40) e θ θ = ( a cos θ a sin θ 0 ) (30.40) e θ θ = a cos θ a sin θ 0 {:(30.40)(dele_(theta))/(del theta)=([-a cos theta],[-a sin theta],[0]):}\frac{\partial \boldsymbol{e}_{\theta}}{\partial \theta}=\left(\begin{array}{c} -a \cos \theta \tag{30.40}\\ -a \sin \theta \\ 0 \end{array}\right)(30.40)eθθ=(acosθasinθ0)
and dotting this with n n n\boldsymbol{n}n yields a a -a-aa. We conclude K θ θ = a K θ θ = a K_(theta theta)=-aK_{\theta \theta}=-aKθθ=a and all other components of K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν vanish.

30.5 Intrinsic and extrinsic curvature

We can now find out how the symmetric matrix K μ ν = d e ν d x μ n ^ K μ ν = d e ν d x μ n ^ K_(mu nu)=(de_(nu))/(dx^(mu))* hat(n)K_{\mu \nu}=\frac{\mathrm{d} \boldsymbol{e}_{\nu}}{\mathrm{d} x^{\mu}} \cdot \hat{\boldsymbol{n}}Kμν=deνdxμn^, that we met in the last section, relates to the Gaussian curvature. Consider a curve lying in the surface that passes through a point P P P\mathcal{P}P (Fig. 30.7), where it has tangent vector t t ttt. As usual, we parametrize the curve with the arc-length parameter s s sss, but we note for later that we also have access to the coordinates x μ x μ x^(mu)x^{\mu}xμ that lie within the surface. Differentiating the tangent vector along the curve, we have
(30.41) d t d s = d t μ d s e μ + t μ d e μ d s (30.41) d t d s = d t μ d s e μ + t μ d e μ d s {:(30.41)(dt)/((d)s)=(dt^(mu))/(ds)e_(mu)+t^(mu)(de_(mu))/(ds):}\begin{equation*} \frac{\mathrm{d} \boldsymbol{t}}{\mathrm{~d} s}=\frac{\mathrm{d} t^{\mu}}{\mathrm{d} s} \boldsymbol{e}_{\mu}+t^{\mu} \frac{\mathrm{d} \boldsymbol{e}_{\mu}}{\mathrm{d} s} \tag{30.41} \end{equation*}(30.41)dt ds=dtμdseμ+tμdeμds
We expand this derivative in terms of two constants, κ g κ g kappa_(g)\kappa_{g}κg and κ n κ n kappa_(n)\kappa_{n}κn, as
(30.42) d t d s = d t μ d s e μ + t μ d e μ d s = κ g e + κ n n ^ (30.42) d t d s = d t μ d s e μ + t μ d e μ d s = κ g e + κ n n ^ {:(30.42)(dt)/((d)s)=(dt^(mu))/(ds)e_(mu)+t^(mu)(de_(mu))/(ds)=kappa_(g)e+kappa_(n) hat(n):}\begin{equation*} \frac{\mathrm{d} t}{\mathrm{~d} s}=\frac{\mathrm{d} t^{\mu}}{\mathrm{d} s} \boldsymbol{e}_{\mu}+t^{\mu} \frac{\mathrm{d} \boldsymbol{e}_{\mu}}{\mathrm{d} s}=\kappa_{\mathrm{g}} \boldsymbol{e}+\kappa_{n} \hat{\boldsymbol{n}} \tag{30.42} \end{equation*}(30.42)dt ds=dtμdseμ+tμdeμds=κge+κnn^
where e e e\boldsymbol{e}e is some unit vector formed from a linear combination of e 1 e 1 e_(1)\boldsymbol{e}_{1}e1 and e 2 e 2 e_(2)\boldsymbol{e}_{2}e2, and is, as such, perpendicular to n ^ n ^ hat(n)\hat{\boldsymbol{n}}n^.
Now consider the derivative d d s d d s (d)/(ds)\frac{\mathrm{d}}{\mathrm{d} s}dds. When represented in terms of the coordinates in the surface, this is
(30.43) d d s = d x ν d s x ν = t ν x ν (30.43) d d s = d x ν d s x ν = t ν x ν {:(30.43)(d)/((d)s)=(dx^(nu))/(ds)*(del)/(delx^(nu))=t^(nu)(del)/(delx^(nu)):}\begin{equation*} \frac{\mathrm{d}}{\mathrm{~d} s}=\frac{\mathrm{d} x^{\nu}}{\mathrm{d} s} \cdot \frac{\partial}{\partial x^{\nu}}=t^{\nu} \frac{\partial}{\partial x^{\nu}} \tag{30.43} \end{equation*}(30.43)d ds=dxνdsxν=tνxν
Therefore, a term t μ d e μ d s t μ d e μ d s t^(mu)(de_(mu))/(ds)t^{\mu} \frac{\mathrm{d} \boldsymbol{e}_{\mu}}{\mathrm{d} s}tμdeμds. can be written as t μ t ν e μ x ν t μ t ν e μ x ν t^(mu)t^(nu)(dele_(mu))/(delx^(nu))t^{\mu} t^{\nu} \frac{\partial \boldsymbol{e}_{\mu}}{\partial x^{\nu}}tμtνeμxν. As a result, when we project eqn 30.42 along n ^ n ^ hat(n)\hat{\boldsymbol{n}}n^ we can extract
(30.44) d t d s n ^ = t μ t ν e μ x ν n ^ = κ n (30.44) d t d s n ^ = t μ t ν e μ x ν n ^ = κ n {:(30.44)(dt)/((d)s)* hat(n)=t^(mu)t^(nu)(dele_(mu))/(delx^(nu))* hat(n)=kappa_(n):}\begin{equation*} \frac{\mathrm{d} \boldsymbol{t}}{\mathrm{~d} s} \cdot \hat{\boldsymbol{n}}=t^{\mu} t^{\nu} \frac{\partial \boldsymbol{e}_{\mu}}{\partial x^{\nu}} \cdot \hat{\boldsymbol{n}}=\kappa_{n} \tag{30.44} \end{equation*}(30.44)dt dsn^=tμtνeμxνn^=κn
where we've used that n ^ e = 0 n ^ e = 0 hat(n)*e=0\hat{\boldsymbol{n}} \cdot \boldsymbol{e}=0n^e=0 and n ^ n ^ = 1 n ^ n ^ = 1 hat(n)* hat(n)=1\hat{\boldsymbol{n}} \cdot \hat{\boldsymbol{n}}=1n^n^=1. Finally, recall from Gauss' equation e μ x ν n ^ = K ν μ = K μ ν e μ x ν n ^ = K ν μ = K μ ν (dele_(mu))/(delx^(nu))* hat(n)=K_(nu mu)=K_(mu nu)\frac{\partial \boldsymbol{e}_{\mu}}{\partial x^{\nu}} \cdot \hat{\boldsymbol{n}}=K_{\nu \mu}=K_{\mu \nu}eμxνn^=Kνμ=Kμν, which says that the part of e μ / x ν e μ / x ν dele_(mu)//delx^(nu)\partial \boldsymbol{e}_{\mu} / \partial x^{\nu}eμ/xν pointing along n ^ n ^ hat(n)\hat{\boldsymbol{n}}n^ is K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν and so 18 18 ^(18){ }^{18}18
(30.47) κ n = t μ t ν K μ ν (30.47) κ n = t μ t ν K μ ν {:(30.47)kappa_(n)=t^(mu)t^(nu)K_(mu nu):}\begin{equation*} \kappa_{n}=t^{\mu} t^{\nu} K_{\mu \nu} \tag{30.47} \end{equation*}(30.47)κn=tμtνKμν
To relate this expression to curvature, we will follow Gauss and look not just at this one curve, but at all of the curves passing through P P P\mathcal{P}P. For each, we calculate κ n κ n kappa_(n)\kappa_{n}κn and then find the extremal value, i.e. the largest and smallest values of κ n κ n kappa_(n)\kappa_{n}κn.
Example 30.9
To do this, we extremize the quantity K μ ν t μ t ν K μ ν t μ t ν K_(mu nu)t^(mu)t^(nu)K_{\mu \nu} t^{\mu} t^{\nu}Kμνtμtν subject to the constraint that the tangent vectors are properly normalized to unity, written 19 19 ^(19){ }^{19}19 as g μ ν t μ t ν = 1 g μ ν t μ t ν = 1 g_(mu nu)t^(mu)t^(nu)=1g_{\mu \nu} t^{\mu} t^{\nu}=1gμνtμtν=1. We can carry out this procedure using a Lagrange multiplier k k kkk. So, following the usual procedure, we calculate
d d t μ [ K μ ν t μ t ν k ( g μ ν t μ t ν 1 ) ] = 0 d d t μ K μ ν t μ t ν k g μ ν t μ t ν 1 = 0 (d)/((d)t^(mu))[K_(mu nu)t^(mu)t^(nu)-k(g_(mu nu)t^(mu)t^(nu)-1)]=0\frac{\mathrm{d}}{\mathrm{~d} t^{\mu}}\left[K_{\mu \nu} t^{\mu} t^{\nu}-k\left(g_{\mu \nu} t^{\mu} t^{\nu}-1\right)\right]=0d dtμ[Kμνtμtνk(gμνtμtν1)]=0
The result is
(0.49) K μ ν t ν k g μ ν t ν = 0 (0.49) K μ ν t ν k g μ ν t ν = 0 {:(0.49)K_(mu nu)t^(nu)-kg_(mu nu)t^(nu)=0:}\begin{equation*} K_{\mu \nu} t^{\nu}-k g_{\mu \nu} t^{\nu}=0 \tag{0.49} \end{equation*}(0.49)Kμνtνkgμνtν=0
Multiply through by the metric components to find
g λ μ K μ ν t ν k δ ν λ t ν = 0 g λ μ K μ ν t ν k δ ν λ t ν = 0 g^(lambda mu)K_(mu nu)t^(nu)-kdelta_(nu)^(lambda)t^(nu)=0g^{\lambda \mu} K_{\mu \nu} t^{\nu}-k \delta_{\nu}^{\lambda} t^{\nu}=0gλμKμνtνkδνλtν=0
This is an eigenvalue equation for the matrix A A _ A_\underline{\boldsymbol{A}}A with components 20 A λ ν = g λ μ K μ ν 20 A λ ν = g λ μ K μ ν ^(20)A^(lambda)_(nu)=g^(lambda mu)K_(mu nu){ }^{20} A^{\lambda}{ }_{\nu}=g^{\lambda \mu} K_{\mu \nu}20Aλν=gλμKμν. It is the eigenvalues k 1 k 1 k_(1)k_{1}k1 and k 2 k 2 k_(2)k_{2}k2 of A A _ A_\underline{\boldsymbol{A}}A that will be important. To see how, we multiply through by t λ = g λ σ t σ t λ = g λ σ t σ t_(lambda)=g_(lambda sigma)t^(sigma)t_{\lambda}=g_{\lambda \sigma} t^{\sigma}tλ=gλσtσ and contract the index λ λ lambda\lambdaλ to obtain
and so
(30.54) t μ K μ ν t ν k t ν t ν = 0 k = K μ ν t μ t ν = κ n extremal (30.54) t μ K μ ν t ν k t ν t ν = 0 k = K μ ν t μ t ν = κ n extremal  {:[(30.54)t^(mu)K_(mu nu)t^(nu)-kt_(nu)t^(nu)=0],[k=K_(mu nu)t^(mu)t^(nu)=kappa_(n)^("extremal ")]:}\begin{gather*} t^{\mu} K_{\mu \nu} t^{\nu}-k t_{\nu} t^{\nu}=0 \tag{30.54}\\ k=K_{\mu \nu} t^{\mu} t^{\nu}=\kappa_{n}^{\text {extremal }} \end{gather*}(30.54)tμKμνtνktνtν=0k=Kμνtμtν=κnextremal 
k = K μ ν t μ t ν = κ n extremal. k = K μ ν t μ t ν = κ n extremal.  k=K_(mu nu)t^(mu)t^(nu)=kappa_(n)^("extremal. ")k=K_{\mu \nu} t^{\mu} t^{\nu}=\kappa_{n}^{\text {extremal. }}k=Kμνtμtν=κnextremal. 
We conclude that the two eigenvalues k 1 k 1 k_(1)k_{1}k1 and k 2 k 2 k_(2)k_{2}k2 of the matrix A A _ A_\underline{\boldsymbol{A}}A with components g λ μ K μ ν g λ μ K μ ν g^(lambda mu)K_(mu nu)g^{\lambda \mu} K_{\mu \nu}gλμKμν give the extremal values of the curvature κ n extremal κ n extremal  kappa_(n)^("extremal ")\kappa_{n}^{\text {extremal }}κnextremal .
The approach from the last example gives us access to two invariants of the matrix A A _ A_\underline{\boldsymbol{A}}A that characterize the curvature: (i) the determinant det A = det K / det g = k 1 k 2 det A _ = det K _ / det g _ = k 1 k 2 detA_=detK_//detg_=k_(1)k_(2)\operatorname{det} \underline{\boldsymbol{A}}=\operatorname{det} \underline{\boldsymbol{K}} / \operatorname{det} \underline{\boldsymbol{g}}=k_{1} k_{2}detA=detK/detg=k1k2 and (ii) the trace Tr A = Tr ( g 1 K ) = Tr A _ = Tr g 1 K _ = TrA_=Tr(g^(-1)K_)=\operatorname{Tr} \underline{\boldsymbol{A}}=\operatorname{Tr}\left(\boldsymbol{g}^{-1} \underline{\boldsymbol{K}}\right)=TrA=Tr(g1K)= k 1 + k ¯ 2 k 1 + k ¯ 2 k_(1)+ bar(k)_(2)k_{1}+\bar{k}_{2}k1+k¯2. It turns out that the Gaussian curvature or intrinsic curvature that we previously called K K KKK is given by the determinant k 1 k 2 k 1 k 2 k_(1)k_(2)k_{1} k_{2}k1k2, while the trace ( k 1 + k 2 ) k 1 + k 2 (k_(1)+k_(2))\left(k_{1}+k_{2}\right)(k1+k2) gives a quantity called the extrinsic curvature. What do these terms mean? A good example is a cylinder (see Fig. 30.8) which has extrinsic curvature by virtue of the way in which the twodimensional surface is embedded in three-dimensional space. However, it is not intrinsically curved because that surface can easily be unwrapped and placed on a flat surface (as shown in the figure).
Unlike the cylinder (see Fig. 30.8), a sphere cannot be made from a flat piece of paper (at least, not without cutting the paper) and so possesses both intrinsic and extrinsic curvature. The cylinder, on the other other hand, has extrinsic curvature but zero intrinsic curvature. It is therefore intrinsic curvature, that cannot be removed by unwrapping the surface without cutting the paper. 21 21 ^(21){ }^{21}21

Example 30.10

In this example, we evaluate the intrinsic and extrinsic curvature of two surfaces embedded in three-dimensional space.
(a) Consider a parabolic surface. This is described by coordinates ( x 1 , x 2 ) = ( x , y ) x 1 , x 2 = ( x , y ) (x^(1),x^(2))=(x,y)\left(x^{1}, x^{2}\right)=(x, y)(x1,x2)=(x,y) with
(30.56) X = ( x y 1 2 u x 2 + 1 2 v y 2 ) (30.56) X = x y 1 2 u x 2 + 1 2 v y 2 {:(30.56)X=([x],[y],[(1)/(2)ux^(2)+(1)/(2)vy^(2)]):}\boldsymbol{X}=\left(\begin{array}{c} x \tag{30.56}\\ y \\ \frac{1}{2} u x^{2}+\frac{1}{2} v y^{2} \end{array}\right)(30.56)X=(xy12ux2+12vy2)
19 19 ^(19){ }^{19}19 Note how the metric is needed here as the surface is, in general, curved.
20 20 ^(20){ }^{20}20 Example: For the cylinder in Example 30.8 we write a matrix with components K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν
(30.51) K = ( a 0 0 0 ) (30.51) K _ = a 0 0 0 {:(30.51)K_=([-a,0],[0,0]):}\underline{K}=\left(\begin{array}{cc} -a & 0 \tag{30.51}\\ 0 & 0 \end{array}\right)(30.51)K=(a000)
and a matrix with components g λ μ g λ μ g^(lambda mu)g^{\lambda \mu}gλμ
(30.52) g = ( 1 / a 2 0 0 1 ) (30.52) g _ = 1 / a 2 0 0 1 {:(30.52)g_=([1//a^(2),0],[0,1]):}\underline{g}=\left(\begin{array}{cc} 1 / a^{2} & 0 \tag{30.52}\\ 0 & 1 \end{array}\right)(30.52)g=(1/a2001)
which gives the matrix
(30.53) A = ( 1 / a 0 0 0 ) (30.53) A _ = 1 / a 0 0 0 {:(30.53)A_=([-1//a,0],[0,0]):}\underline{\boldsymbol{A}}=\left(\begin{array}{cc} -1 / a & 0 \tag{30.53}\\ 0 & 0 \end{array}\right)(30.53)A=(1/a000)
Fig. 30.8 A cylinder (the tin can) looks curved, but it is not intrinsically curved but only curved extrinsically (by the way it is embedded in three-dimensional space). The cylindrical surface can be unwound and placed cal surface can be unwound and placed
on a flat surface without tearing or dison a flat surface without tearing or dis-
torting. For a cylinder of radius a a aaa, the eigenvalues of the matrix A A _ A_\underline{\boldsymbol{A}}A are k 1 = 1 / a k 1 = 1 / a k_(1)=-1//ak_{1}=-1 / ak1=1/a and k 2 = 0 k 2 = 0 k_(2)=0k_{2}=0k2=0, so we have zero intrinsic curvature K = k 1 k 2 = 0 K = k 1 k 2 = 0 K=k_(1)k_(2)=0K=k_{1} k_{2}=0K=k1k2=0, even though the extrinsic curvature is non-zero ( k 1 + k 2 = 1 / a k 1 + k 2 = 1 / a k_(1)+k_(2)=-1//ak_{1}+k_{2}=-1 / ak1+k2=1/a ).
21 21 ^(21){ }^{21}21 For one-dimensional curves, a notion of intrinsic curvature doesn't really make sense, since the curve can always be flattened out; however curved around your shoelaces are, you can always unthread them and lay them out in straight lines. Thus, onedimensional curves only possess extrinsic curvature.
22 22 ^(22){ }^{22}22 Recall that the metric tensor g g g\boldsymbol{g}g has components
(30.60) g μ ν = ( a 2 0 0 a 2 sin 2 θ ) (30.60) g μ ν = a 2 0 0 a 2 sin 2 θ {:(30.60)g_(mu nu)=([a^(2),0],[0,a^(2)sin^(2)theta]):}g_{\mu \nu}=\left(\begin{array}{cc} a^{2} & 0 \tag{30.60}\\ 0 & a^{2} \sin ^{2} \theta \end{array}\right)(30.60)gμν=(a200a2sin2θ) so that det g = a 4 sin 2 θ det g = a 4 sin 2 θ det g=a^(4)sin^(2)theta\operatorname{det} \boldsymbol{g}=a^{4} \sin ^{2} \thetadetg=a4sin2θ. The inverse metric has components
(30.61) g μ ν = ( 1 a 2 0 0 1 a 2 sin 2 θ ) (30.61) g μ ν = 1 a 2 0 0 1 a 2 sin 2 θ {:(30.61)g^(mu nu)=([(1)/(a^(2)),0],[0,(1)/(a^(2)sin^(2)theta)]):}g^{\mu \nu}=\left(\begin{array}{cc} \frac{1}{a^{2}} & 0 \tag{30.61}\\ 0 & \frac{1}{a^{2} \sin ^{2} \theta} \end{array}\right)(30.61)gμν=(1a2001a2sin2θ)
23 23 ^(23){ }^{23}23 Riemann sets out his aims in his inaugural lecture delivered at the University of Göttingen in 1853. The lecture was intended to be accessible to the entire faculty and so contains minimal mathematical detail. It's immense significance is explained in Spivak, Vol. II
24 24 ^(24){ }^{24}24 Why? Since g μ ν = g ν μ g μ ν = g ν μ g_(mu nu)=g_(nu mu)g_{\mu \nu}=g_{\nu \mu}gμν=gνμ, then the independent components (i) are the diagonal elements (of which there are n n nnn ) and (ii) half of the off-diagonal el ements (and there are ( n 2 n ) n 2 n (n^(2)-n)\left(n^{2}-n\right)(n2n) offdiagonal elements). Adding these we have n + ( n 2 n ) / 2 = n ( n + 1 ) / 2 n + n 2 n / 2 = n ( n + 1 ) / 2 n+(n^(2)-n)//2=n(n+1)//2n+\left(n^{2}-n\right) / 2=n(n+1) / 2n+(n2n)/2=n(n+1)/2, as claimed.
25 25 ^(25){ }^{25}25 The function Q Q QQQ is a function of 2 n 2 n 2n2 n2n variables which take the form of two vectors X X X\boldsymbol{X}X and Y Y Y\boldsymbol{Y}Y. We assume these two vectors span the two-dimensional space W W WWW and simply write Q ( W ) Q ( W ) Q(W)Q(W)Q(W) for simplicity here.
The metric is then described by the matrix
(30.57) g μ ν = ( 1 + u 2 x 2 u v x y u v x y 1 + v 2 y 2 ) (30.57) g μ ν = 1 + u 2 x 2 u v x y u v x y 1 + v 2 y 2 {:(30.57)g_(mu nu)=([1+u^(2)x^(2),uvxy],[uvxy,1+v^(2)y^(2)]):}g_{\mu \nu}=\left(\begin{array}{cc} 1+u^{2} x^{2} & u v x y \tag{30.57}\\ u v x y & 1+v^{2} y^{2} \end{array}\right)(30.57)gμν=(1+u2x2uvxyuvxy1+v2y2)
We have the basis vectors and unit normal
(30.58) e x = ( 1 0 u x ) , e y = ( 0 1 v y ) , n ^ = 1 | n | ( u x v y 1 ) (30.58) e x = 1 0 u x , e y = 0 1 v y , n ^ = 1 | n | u x v y 1 {:(30.58)e_(x)=([1],[0],[ux])","quade_(y)=([0],[1],[vy])","quad hat(n)=(1)/(|n|)([-ux],[-vy],[1]):}\boldsymbol{e}_{x}=\left(\begin{array}{c} 1 \tag{30.58}\\ 0 \\ u x \end{array}\right), \quad \boldsymbol{e}_{y}=\left(\begin{array}{c} 0 \\ 1 \\ v y \end{array}\right), \quad \hat{\boldsymbol{n}}=\frac{1}{|\boldsymbol{n}|}\left(\begin{array}{c} -u x \\ -v y \\ 1 \end{array}\right)(30.58)ex=(10ux),ey=(01vy),n^=1|n|(uxvy1)
where | n | = ( 1 + u 2 x 2 + v 2 y 2 ) 1 2 | n | = 1 + u 2 x 2 + v 2 y 2 1 2 |n|=(1+u^(2)x^(2)+v^(2)y^(2))^((1)/(2))|\boldsymbol{n}|=\left(1+u^{2} x^{2}+v^{2} y^{2}\right)^{\frac{1}{2}}|n|=(1+u2x2+v2y2)12. Taking derivatives yields
(30.59) e x x = ( 0 0 u ) , e y y = ( 0 0 v ) (30.59) e x x = 0 0 u , e y y = 0 0 v {:(30.59)(dele_(x))/(del x)=([0],[0],[u])","quad(dele_(y))/(del y)=([0],[0],[v]):}\frac{\partial \boldsymbol{e}_{x}}{\partial x}=\left(\begin{array}{l} 0 \tag{30.59}\\ 0 \\ u \end{array}\right), \quad \frac{\partial \boldsymbol{e}_{y}}{\partial y}=\left(\begin{array}{l} 0 \\ 0 \\ v \end{array}\right)(30.59)exx=(00u),eyy=(00v)
with other derivatives vanishing. Let's concentrate on the curvature evaluated at the origin, where we have that g u u = g v v = 1 , g u v = g v u = 0 , K 11 = u g u u = g v v = 1 , g u v = g v u = 0 , K 11 = u g_(uu)=g_(vv)=1,g_(uv)=g_(vu)=0,K_(11)=ug_{u u}=g_{v v}=1, g_{u v}=g_{v u}=0, K_{11}=uguu=gvv=1,guv=gvu=0,K11=u and K 22 = v K 22 = v K_(22)=vK_{22}=vK22=v. We have that the intrinsic curvature is K = u v K = u v K=uvK=u vK=uv and the extrinsic curvature is u + v u + v u+vu+vu+v. (b) Next, consider the spherical surface with coordinates ( x 1 , x 2 ) = ( θ , ϕ ) x 1 , x 2 = ( θ , ϕ ) (x^(1),x^(2))=(theta,phi)\left(x^{1}, x^{2}\right)=(\theta, \phi)(x1,x2)=(θ,ϕ). The surface is written as 22 22 ^(22){ }^{22}22
(30.62) X = ( a sin θ cos ϕ a sin θ sin ϕ a cos θ ) (30.62) X = a sin θ cos ϕ a sin θ sin ϕ a cos θ {:(30.62)X=([a sin theta cos phi],[a sin theta sin phi],[a cos theta]):}\boldsymbol{X}=\left(\begin{array}{c} a \sin \theta \cos \phi \tag{30.62}\\ a \sin \theta \sin \phi \\ a \cos \theta \end{array}\right)(30.62)X=(asinθcosϕasinθsinϕacosθ)
The basis vectors and unit normal follow as
e θ = ( a cos θ cos ϕ a cos θ sin ϕ a sin θ ) , e ϕ = ( a sin θ sin ϕ a sin θ cos ϕ 0 ) , n ^ = ( sin θ cos ϕ sin θ sin ϕ cos θ ) e θ = a cos θ cos ϕ a cos θ sin ϕ a sin θ , e ϕ = a sin θ sin ϕ a sin θ cos ϕ 0 , n ^ = sin θ cos ϕ sin θ sin ϕ cos θ e_(theta)=([a cos theta cos phi],[a cos theta sin phi],[-a sin theta]),quade_(phi)=([-a sin theta sin phi],[a sin theta cos phi],[0]),quad hat(n)=([sin theta cos phi],[sin theta sin phi],[cos theta])\boldsymbol{e}_{\theta}=\left(\begin{array}{c} a \cos \theta \cos \phi \\ a \cos \theta \sin \phi \\ -a \sin \theta \end{array}\right), \quad \boldsymbol{e}_{\phi}=\left(\begin{array}{c} -a \sin \theta \sin \phi \\ a \sin \theta \cos \phi \\ 0 \end{array}\right), \quad \hat{\boldsymbol{n}}=\left(\begin{array}{c} \sin \theta \cos \phi \\ \sin \theta \sin \phi \\ \cos \theta \end{array}\right)eθ=(acosθcosϕacosθsinϕasinθ),eϕ=(asinθsinϕasinθcosϕ0),n^=(sinθcosϕsinθsinϕcosθ)
Taking derivatives we obtain
e θ θ = a n ^ , e θ ϕ = e ϕ θ = a cot θ e ϕ , e ϕ ϕ = a sin θ cos θ e 1 a sin 2 θ n ^ e θ θ = a n ^ , e θ ϕ = e ϕ θ = a cot θ e ϕ , e ϕ ϕ = a sin θ cos θ e 1 a sin 2 θ n ^ (dele_(theta))/(del theta)=-a hat(n),quad(dele_(theta))/(del phi)=(dele_(phi))/(del theta)=a cot thetae_(phi),quad(dele_(phi))/(del phi)=-a sin theta cos thetae_(1)-asin^(2)theta hat(n)\frac{\partial \boldsymbol{e}_{\theta}}{\partial \theta}=-a \hat{\boldsymbol{n}}, \quad \frac{\partial \boldsymbol{e}_{\theta}}{\partial \phi}=\frac{\partial \boldsymbol{e}_{\phi}}{\partial \theta}=a \cot \theta \boldsymbol{e}_{\phi}, \quad \frac{\partial \boldsymbol{e}_{\phi}}{\partial \phi}=-a \sin \theta \cos \theta \boldsymbol{e}_{1}-a \sin ^{2} \theta \hat{\boldsymbol{n}}eθθ=an^,eθϕ=eϕθ=acotθeϕ,eϕϕ=asinθcosθe1asin2θn^.
(30.64)
We have non-zero components K 11 = a K 11 = a K_(11)=-aK_{11}=-aK11=a and K 22 = a sin 2 θ K 22 = a sin 2 θ K_(22)=-asin^(2)thetaK_{22}=-a \sin ^{2} \thetaK22=asin2θ. We find an intrinsic curvature det K / det g = 1 / a 2 det K _ / det g _ = 1 / a 2 detK_//detg_=1//a^(2)\operatorname{det} \underline{\boldsymbol{K}} / \operatorname{det} \underline{\boldsymbol{g}}=1 / a^{2}detK/detg=1/a2 and an extrinsic curvature Tr ( g 1 K ) = 2 / a Tr g _ 1 K _ = 2 / a Tr(g_^(-1)K_)=-2//a\operatorname{Tr}\left(\underline{\boldsymbol{g}}^{-1} \underline{\boldsymbol{K}}\right)=-2 / aTr(g1K)=2/a.

30.6 Riemann's project

Bernhard Riemann aimed to generalize Gauss' remarkable result so that any observer confined to an n n nnn-dimensional space would be able to evaluate the intrinsic curvature. 23 23 ^(23){ }^{23}23 The success of his approach laid the foundations for general relativity.
Broadly speaking, Riemann's method was to ask: when does the introduction of a new coordinate system change a metric g g g\boldsymbol{g}g with components g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν, defined in n n nnn-dimensional space, into some other metric a a a\boldsymbol{a}a with components a μ ν a μ ν a_(mu nu)a_{\mu \nu}aμν ? Riemann argued that the metric g g g\boldsymbol{g}g is determined by n ( n + 1 ) / 2 n ( n + 1 ) / 2 n(n+1)//2n(n+1) / 2n(n+1)/2 functions 24 24 ^(24){ }^{24}24 but that a new coordinate system can be defined using n n nnn functions. As a result, the new metric must be determined by n ( n + 1 ) / 2 n = n ( n 1 ) / 2 n ( n + 1 ) / 2 n = n ( n 1 ) / 2 n(n+1)//2-n=n(n-1)//2n(n+1) / 2-n=n(n-1) / 2n(n+1)/2n=n(n1)/2 functions. Riemann then claimed that there is a quadratic function 25 Q 25 Q ^(25)Q{ }^{25} Q25Q that uniquely assigns a number to a two-dimensional space W W WWW and that (i) for a two-dimensional manifold 3 Q ( W ) 3 Q ( W ) -3Q(W)-3 Q(W)3Q(W) is the Gaussian curvature; (ii) that for higher dimensional manifolds, 3 Q ( W ) 3 Q ( W ) -3Q(W)-3 Q(W)3Q(W) describes the Gaussian curvature of the two-dimensional subspace W W WWW of the manifold; and (iii) there are n ( n 1 ) / 2 n ( n 1 ) / 2 n(n-1)//2n(n-1) / 2n(n1)/2 independent
two-dimensional subspaces for an n n nnn-dimensional vector space; if Q ( W ) Q ( W ) Q(W)Q(W)Q(W) is known for each of these, then the metric is completely determined. This all seems a little obscure, but was shortly to be reformulated into something more familiar.
Some eight years later, in 1861, Riemann submitted a paper to the Paris Academy as an entry to a competition aiming to provide the answer to a question in the problem of heat conduction. 26 26 ^(26){ }^{26}26 The paper contains the first instance of (what we would now call) the Riemann curvature tensor R ( , R ( , R(,\boldsymbol{R}(,R(,, , ) in terms of the components of the metric. Riemann's paper asked what conditions make a space flat and gave the answer that what is needed is that the components of R R R\boldsymbol{R}R should vanish. The number Q ( W ) Q ( W ) Q(W)Q(W)Q(W) turns out to be proportional to the output of the ( 0 , 4 ) ( 0 , 4 ) (0,4)(0,4)(0,4) version of the tensor R R R\boldsymbol{R}R when its slots are filled by linearly independent vectors X X X\boldsymbol{X}X and Y Y Y\boldsymbol{Y}Y that span the two-dimensional subspace W . 27 W . 27 W.^(27)W .{ }^{27}W.27 The significance of this is that the curvature of space determines the metric.
We shall not examine the details of Riemann's method in any more detail. 28 28 ^(28){ }^{28}28 The important point is that Riemann's curvature tensor, which can be written in terms of the components of the metric only, not only contains the Gaussian curvature as a special case in two dimensions but generalizes a notion of curvature to higher dimensional cases. The promised link in two dimensions between the Gaussian curvature K K KKK and the Riemann tensor is
(30.65) K = R 1212 g , (30.65) K = R 1212 g {:(30.65)K=(R_(1212))/(g)", ":}\begin{equation*} K=\frac{R_{1212}}{g} \text {, } \tag{30.65} \end{equation*}(30.65)K=R1212g
where g = det g μ ν g = det g μ ν g=detg_(mu nu)g=\operatorname{det} g_{\mu \nu}g=detgμν, the determinant of the metric tensor g g g\boldsymbol{g}g.
Example 30.11
We can now justify the link between the Riemann tensor and the Gaussian curvature. The simplest route is to use eqn 30.65 to write 4 g R 1212 = 4 g 2 K 4 g R 1212 = 4 g 2 K 4gR_(1212)=4g^(2)K4 g R_{1212}=4 g^{2} K4gR1212=4g2K. We then make use of the relationship from eqn 11.22 between the coordinates of R R R\boldsymbol{R}R and the metric
(30.66) R 1212 = 1 2 ( g 12 , 21 g 11 , 22 + g 21 , 12 g 22 , 11 ) + g σ ρ ( Γ 21 σ Γ 12 ρ Γ 22 σ Γ 11 ρ ) . (30.66) R 1212 = 1 2 g 12 , 21 g 11 , 22 + g 21 , 12 g 22 , 11 + g σ ρ Γ 21 σ Γ 12 ρ Γ 22 σ Γ 11 ρ . {:(30.66)R_(1212)=(1)/(2)(g_(12,21)-g_(11,22)+g_(21,12)-g_(22,11))+g_(sigma rho)(Gamma_(21)^(sigma)Gamma_(12)^(rho)-Gamma_(22)^(sigma)Gamma_(11)^(rho)).:}\begin{equation*} R_{1212}=\frac{1}{2}\left(g_{12,21}-g_{11,22}+g_{21,12}-g_{22,11}\right)+g_{\sigma \rho}\left(\Gamma_{21}^{\sigma} \Gamma_{12}^{\rho}-\Gamma_{22}^{\sigma} \Gamma_{11}^{\rho}\right) . \tag{30.66} \end{equation*}(30.66)R1212=12(g12,21g11,22+g21,12g22,11)+gσρ(Γ21σΓ12ρΓ22σΓ11ρ).
We also need the expression for the connection coefficients
(30.67) Γ μ σ ρ = g ρ λ 2 ( g λ μ x σ + g λ σ x μ g μ σ x λ ) (30.67) Γ μ σ ρ = g ρ λ 2 g λ μ x σ + g λ σ x μ g μ σ x λ {:(30.67)Gamma_(mu sigma)^(rho)=(g^(rho lambda))/(2)((delg_(lambda mu))/(delx^(sigma))+(delg_(lambda sigma))/(delx^(mu))-(delg_(mu sigma))/(delx^(lambda))):}\begin{equation*} \Gamma_{\mu \sigma}^{\rho}=\frac{g^{\rho \lambda}}{2}\left(\frac{\partial g_{\lambda \mu}}{\partial x^{\sigma}}+\frac{\partial g_{\lambda \sigma}}{\partial x^{\mu}}-\frac{\partial g_{\mu \sigma}}{\partial x^{\lambda}}\right) \tag{30.67} \end{equation*}(30.67)Γμσρ=gρλ2(gλμxσ+gλσxμgμσxλ)
so that, for example,
(30.68) Γ 12 ρ = g ρ 1 ( g 11 , 2 + g 12 , 1 g 12 , 1 ) + g ρ 2 ( g 21 , 1 + g 22 , 1 g 12 , 2 ) (30.68) Γ 12 ρ = g ρ 1 g 11 , 2 + g 12 , 1 g 12 , 1 + g ρ 2 g 21 , 1 + g 22 , 1 g 12 , 2 {:(30.68)Gamma_(12)^(rho)=g^(rho1)(g_(11,2)+g_(12,1)-g_(12,1))+g^(rho2)(g_(21,1)+g_(22,1)-g_(12,2)):}\begin{equation*} \Gamma_{12}^{\rho}=g^{\rho 1}\left(g_{11,2}+g_{12,1}-g_{12,1}\right)+g^{\rho 2}\left(g_{21,1}+g_{22,1}-g_{12,2}\right) \tag{30.68} \end{equation*}(30.68)Γ12ρ=gρ1(g11,2+g12,1g12,1)+gρ2(g21,1+g22,1g12,2)
If we also use the result that
(30.69) g 11 = g 22 / g , g 12 = g 21 = g 12 / g = g 21 / g , g 22 = g 22 / g (30.69) g 11 = g 22 / g , g 12 = g 21 = g 12 / g = g 21 / g , g 22 = g 22 / g {:(30.69)g^(11)=g_(22)//g","quadg^(12)=g^(21)=-g_(12)//g=g_(21)//g","quadg^(22)=g_(22)//g:}\begin{equation*} g^{11}=g_{22} / g, \quad g^{12}=g^{21}=-g_{12} / g=g_{21} / g, \quad g^{22}=g_{22} / g \tag{30.69} \end{equation*}(30.69)g11=g22/g,g12=g21=g12/g=g21/g,g22=g22/g
then, put all together with eqn 30.25 allows one to verify the claim. Try it and see!
This completes our review of classical curvature. In the next chapter, we shall begin to examine the more modern approach to geometry that most naturally fits with the physics of general relativity.
26 26 ^(26){ }^{26}26 Riemann, whose health was poor at the time, didn't win the prize owing to the lack of detail in his paper. Nobody else won the prize and it was withdrawn in 1868.
27 27 ^(27){ }^{27}27 The explicit equation is 3 Q ( X , Y ) = 3 Q ( X , Y ) = 3Q(X,Y)=3 Q(\boldsymbol{X}, \boldsymbol{Y})=3Q(X,Y)= R ( X , Y , X , Y ) R ( X , Y , X , Y ) -R(X,Y,X,Y)-\boldsymbol{R}(\boldsymbol{X}, \boldsymbol{Y}, \boldsymbol{X}, \boldsymbol{Y})R(X,Y,X,Y), where R R R\boldsymbol{R}R is the ( 0 , 4 ) ( 0 , 4 ) (0,4)(0,4)(0,4) version of the Riemann tensor, with components R μ ν α β R μ ν α β R_(mu nu alpha beta)R_{\mu \nu \alpha \beta}Rμναβ.
28 28 ^(28){ }^{28}28 The interested reader should consult the masterful discussion in Spivak, Vol. the 1.

Chapter summary

  • The curvature of a line can be described in differential geometry using the Frenet-Serret equations.
  • Gaussian curvature allows the measurement of the intrinsic curvature of a two-dimensional surface, using measurements made entirely within that surface.
  • Riemann's result upgrades the Gaussian curvature to higher dimensions.

Exercises

(30.1) A mechanical definition of the curvature of a curve is possible if we imagine a unit mass traversing the curve (confined to a plane) at unit speed. The curvature is the force perpendicular to the curve required to maintain this motion. Prove that this definition is equivalent to that used in this chapter.
(30.2) Another method to compute Gaussian curvature is to use parallel transport to carry a vector around a loop on a surface. We then define
(30.70) ( Gaussian curvature ) = ( angle vector turns through ) ( area of loop ) (30.70) (  Gaussian   curvature  ) = (  angle vector   turns through  ) (  area   of loop  ) {:(30.70)((" Gaussian ")/(" curvature "))=(((" angle vector ")/(" turns through ")))/(((" area ")/(" of loop "))):}\begin{equation*} \binom{\text { Gaussian }}{\text { curvature }}=\frac{\binom{\text { angle vector }}{\text { turns through }}}{\binom{\text { area }}{\text { of loop }}} \tag{30.70} \end{equation*}(30.70)( Gaussian  curvature )=( angle vector  turns through )( area  of loop )
Show that for a loop on the surface of a 2-sphere
(30.71) ( curvature ) = 1 a 2 (30.71) (  curvature  ) = 1 a 2 {:(30.71)(" curvature ")=(1)/(a^(2)):}\begin{equation*} (\text { curvature })=\frac{1}{a^{2}} \tag{30.71} \end{equation*}(30.71)( curvature )=1a2
(30.3) We first met the Bertand-Diquet-Puiseux theorem in Exercise 3.3. Here we use it again.
Another way of describing Gaussian curvature is to use the (normalized) difference between the circumference of a circle in the plane and a circle on the surface. The circumference of the circular locus of points a distance ϵ ϵ epsilon\epsilonϵ from a point differs from the Euclidean value 2 π ϵ 2 π ϵ 2pi epsilon2 \pi \epsilon2πϵ by a correction factor. The curvature is defined as
(30.72) K = lim ϵ 0 6 ϵ 2 ( 1 circumference 2 π ϵ ) (30.72) K = lim ϵ 0 6 ϵ 2 1  circumference  2 π ϵ {:(30.72)K=lim_(epsilon rarr0)(6)/(epsilon^(2))*(1-(" circumference ")/(2pi epsilon)):}\begin{equation*} K=\lim _{\epsilon \rightarrow 0} \frac{6}{\epsilon^{2}} \cdot\left(1-\frac{\text { circumference }}{2 \pi \epsilon}\right) \tag{30.72} \end{equation*}(30.72)K=limϵ06ϵ2(1 circumference 2πϵ)
Use this theorem to work out whether a space described by the following metric is flat
(30.73) d s 2 = d r 2 + sin 2 r d θ 2 (30.73) d s 2 = d r 2 + sin 2 r d θ 2 {:(30.73)ds^(2)=dr^(2)+sin^(2)rdtheta^(2):}\begin{equation*} \mathrm{d} s^{2}=\mathrm{d} r^{2}+\sin ^{2} r \mathrm{~d} \theta^{2} \tag{30.73} \end{equation*}(30.73)ds2=dr2+sin2r dθ2
(30.4) (a) Consider a matrix form of eqn 30.49 . Show that
(30.74) det ( g 1 K k I ) = 0 (30.74) det g _ 1 K _ k I _ = 0 {:(30.74)det(g_^(-1)K_-kI_)=0:}\begin{equation*} \operatorname{det}\left(\underline{g}^{-1} \underline{\boldsymbol{K}}-k \underline{\boldsymbol{I}}\right)=0 \tag{30.74} \end{equation*}(30.74)det(g1KkI)=0
(b) By expanding the previous equation in terms of a trace, show that
(30.75) k 2 Tr ( g 1 K ) k + det ( g 1 K ) = 0 (30.75) k 2 Tr g _ 1 K _ k + det g _ 1 K _ = 0 {:(30.75)k^(2)-Tr(g_^(-1)K_)k+det(g_^(-1)K_)=0:}\begin{equation*} k^{2}-\operatorname{Tr}\left(\underline{\boldsymbol{g}}^{-1} \underline{\boldsymbol{K}}\right) k+\operatorname{det}\left(\underline{\boldsymbol{g}}^{-1} \underline{\boldsymbol{K}}\right)=0 \tag{30.75} \end{equation*}(30.75)k2Tr(g1K)k+det(g1K)=0
(c) Use this to argue that
(30.76) det ( g 1 K ) = det K det g (30.76) det g _ 1 K _ = det K _ det g _ {:(30.76)det(g_^(-1)K_)=(detK_)/(detg_):}\begin{equation*} \operatorname{det}\left(\underline{g}^{-1} \underline{\boldsymbol{K}}\right)=\frac{\operatorname{det} \underline{\boldsymbol{K}}}{\operatorname{det} \underline{\boldsymbol{g}}} \tag{30.76} \end{equation*}(30.76)det(g1K)=detKdetg
gives the product of eigenvalues, and
(30.77) Tr ( g 1 K ) (30.77) Tr g _ 1 K _ {:(30.77)Tr(g_^(-1)K_):}\begin{equation*} \operatorname{Tr}\left(\underline{g}^{-1} \underline{\boldsymbol{K}}\right) \tag{30.77} \end{equation*}(30.77)Tr(g1K)
represents the sum of the eigenvalues.
(30.5) (a) Consider a curve embedded in a surface. Differentiate the equation t n ^ = 0 t n ^ = 0 t* hat(n)=0\boldsymbol{t} \cdot \hat{\boldsymbol{n}}=0tn^=0 along the curve and show that
(30.78) t d n ^ d s = n ^ d t d s = K μ ν t μ t ν (30.78) t d n ^ d s = n ^ d t d s = K μ ν t μ t ν {:(30.78)t*(d( hat(n)))/((d)s)=- hat(n)*(dt)/((d)s)=-K_(mu nu)t^(mu)t^(nu):}\begin{equation*} \boldsymbol{t} \cdot \frac{\mathrm{d} \hat{\boldsymbol{n}}}{\mathrm{~d} s}=-\hat{\boldsymbol{n}} \cdot \frac{\mathrm{d} \boldsymbol{t}}{\mathrm{~d} s}=-K_{\mu \nu} t^{\mu} t^{\nu} \tag{30.78} \end{equation*}(30.78)tdn^ ds=n^dt ds=Kμνtμtν
(b) The previous equation can be used to derive a relationship between n ^ n ^ hat(n)\hat{\boldsymbol{n}}n^ and K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν. Show that
(30.79) g μ σ n ^ σ x ν + K μ ν = 0 (30.79) g μ σ n ^ σ x ν + K μ ν = 0 {:(30.79)g_(mu sigma)(del hat(n)^(sigma))/(delx^(nu))+K_(mu nu)=0:}\begin{equation*} g_{\mu \sigma} \frac{\partial \hat{n}^{\sigma}}{\partial x^{\nu}}+K_{\mu \nu}=0 \tag{30.79} \end{equation*}(30.79)gμσn^σxν+Kμν=0
This is known as Weingarten's equation, which is usually written as
(30.80) K μ ν = e μ n ^ x ν (30.80) K μ ν = e μ n ^ x ν {:(30.80)K_(mu nu)=-e_(mu)*(del( hat(n)))/(delx^(nu)):}\begin{equation*} K_{\mu \nu}=-\boldsymbol{e}_{\mu} \cdot \frac{\partial \hat{\boldsymbol{n}}}{\partial x^{\nu}} \tag{30.80} \end{equation*}(30.80)Kμν=eμn^xν
and gives us another method of extracting K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν. (c) Using Weingarten's equation in the form
(30.81) n ^ x ν = K ν μ e μ , (30.81) n ^ x ν = K ν μ e μ , {:(30.81)(del( hat(n)))/(delx^(nu))=-K_(nu)^(mu)e_(mu)",":}\begin{equation*} \frac{\partial \hat{\boldsymbol{n}}}{\partial x^{\nu}}=-K_{\nu}^{\mu} \boldsymbol{e}_{\mu}, \tag{30.81} \end{equation*}(30.81)n^xν=Kνμeμ,
show
(30.82) K = det K det g (30.82) K = det K _ det g {:(30.82)K=(detK_)/(det g):}\begin{equation*} K=\frac{\operatorname{det} \underline{\boldsymbol{K}}}{\operatorname{det} \boldsymbol{g}} \tag{30.82} \end{equation*}(30.82)K=detKdetg
as we had before.
(30.6) (a) Verify the result for the parabolic surface in Example 30.10 using Gauss's equation (eqn 30.37) to compute the components K μ ν K μ ν K_(mu nu)K_{\mu \nu}Kμν for a general point (i.e. not the origin).
(b) Show that the same result is found using Weingarten's equation from the previous problem.
(c) Verify the result is obtained for the intrinsic curvature K K KKK using eqn 30.35 .
(30.7) Consider a three-dimensional spacelike hypersurface Σ Σ Sigma\SigmaΣ with a timelike unit-normal n n n\boldsymbol{n}n. Construct a ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) projection operator P P P\boldsymbol{P}P with components
(30.83) P ν μ = δ ν μ + n μ n ν . (30.83) P ν μ = δ ν μ + n μ n ν . {:(30.83)P_(nu)^(mu)=delta_(nu)^(mu)+n^(mu)n_(nu).:}\begin{equation*} P_{\nu}^{\mu}=\delta_{\nu}^{\mu}+n^{\mu} n_{\nu} . \tag{30.83} \end{equation*}(30.83)Pνμ=δνμ+nμnν.
(a) Evaluate P μ α P α ν P μ α P α ν P^(mu)_(alpha)P^(alpha)_(nu)P^{\mu}{ }_{\alpha} P^{\alpha}{ }_{\nu}PμαPαν.
(b) Show that the operator P ν μ P ν μ P_(nu)^(mu)P_{\nu}^{\mu}Pνμ acts on vector X X X\boldsymbol{X}X to project a vector X X X\boldsymbol{X}X that is tangent to the surface.
(c) Define an induced metric h h hhh via
(30.84) h α β = P α μ P β ν g μ ν (30.84) h α β = P α μ P β ν g μ ν {:(30.84)h_(alpha beta)=P_(alpha)^(mu)P_(beta)^(nu)g_(mu nu):}\begin{equation*} h_{\alpha \beta}=P_{\alpha}^{\mu} P_{\beta}^{\nu} g_{\mu \nu} \tag{30.84} \end{equation*}(30.84)hαβ=PαμPβνgμν
Show that (i) h α β = g α β + n α n β h α β = g α β + n α n β h_(alpha beta)=g_(alpha beta)+n_(alpha)n_(beta)h_{\alpha \beta}=g_{\alpha \beta}+n_{\alpha} n_{\beta}hαβ=gαβ+nαnβ and (ii)
h α β X ¯ α Y ¯ β = g α β X ¯ α Y ¯ β h α β X ¯ α Y ¯ β = g α β X ¯ α Y ¯ β h_(alpha beta) bar(X)^(alpha) bar(Y)^(beta)=g_(alpha beta) bar(X)^(alpha) bar(Y)^(beta)h_{\alpha \beta} \bar{X}^{\alpha} \bar{Y}^{\beta}=g_{\alpha \beta} \bar{X}^{\alpha} \bar{Y}^{\beta}hαβX¯αY¯β=gαβX¯αY¯β.
(30.8) Now define an induced covariant derivative D D DDD, whose components are given via
(30.85) ( D μ X ) ν = P α ν P μ β ( β X ) α , (30.85) D μ X ν = P α ν P μ β β X α , {:(30.85)(D_(mu)X)^(nu)=P_(alpha)^(nu)P_(mu)^(beta)(grad_(beta)X)^(alpha)",":}\begin{equation*} \left(\boldsymbol{D}_{\mu} \boldsymbol{X}\right)^{\nu}=P_{\alpha}^{\nu} P_{\mu}^{\beta}\left(\boldsymbol{\nabla}_{\beta} \boldsymbol{X}\right)^{\alpha}, \tag{30.85} \end{equation*}(30.85)(DμX)ν=PανPμβ(βX)α,
where, now, the vector field X X X\boldsymbol{X}X is restricted to be tangent to Σ Σ Sigma\SigmaΣ. The double projection, as opposed to the single projection P β μ ( β X ) α P β μ β X α P^(beta)_(mu)(grad_(beta)X)^(alpha)P^{\beta}{ }_{\mu}\left(\boldsymbol{\nabla}_{\beta} \boldsymbol{X}\right)^{\alpha}Pβμ(βX)α, is designed to output a derivative as a vector that is within Σ Σ Sigma\SigmaΣ, cutting off the normal part.
(a) By generalizing this definition, show that ( D μ h ) α β = 0 D μ h α β = 0 (D_(mu)h)_(alpha beta)=0\left(\boldsymbol{D}_{\mu} \boldsymbol{h}\right)_{\alpha \beta}=0(Dμh)αβ=0.
(b) Show that, since X X X\boldsymbol{X}X is tangent to Σ Σ Sigma\SigmaΣ, we have P ν μ X ν = X μ P ν μ X ν = X μ P_(nu)^(mu)X^(nu)=X^(mu)P_{\nu}^{\mu} X^{\nu}=X^{\mu}PνμXν=Xμ.
(c) Starting from the fact that n μ X μ = 0 n μ X μ = 0 n_(mu)X^(mu)=0n_{\mu} X^{\mu}=0nμXμ=0, show further that
(30.86) n β P α μ ( μ X ) β = X ν P ν β P α μ ( μ n ) β (30.86) n β P α μ μ X β = X ν P ν β P α μ μ n β {:(30.86)n_(beta)P_(alpha)^(mu)(grad_(mu)X)^(beta)=-X^(nu)P_(nu)^(beta)P_(alpha)^(mu)(grad_(mu)n)_(beta):}\begin{equation*} n_{\beta} P_{\alpha}^{\mu}\left(\boldsymbol{\nabla}_{\mu} \boldsymbol{X}\right)^{\beta}=-X^{\nu} P_{\nu}^{\beta} P_{\alpha}^{\mu}\left(\boldsymbol{\nabla}_{\mu} \boldsymbol{n}\right)_{\beta} \tag{30.86} \end{equation*}(30.86)nβPαμ(μX)β=XνPνβPαμ(μn)β
This motivates the definition of the extrinsic curvature tensor K K K\boldsymbol{K}K with components
(30.87) K α β = P α μ P β ν ( μ n ) ν (30.87) K α β = P α μ P β ν μ n ν {:(30.87)K_(alpha beta)=-P_(alpha)^(mu)P_(beta)^(nu)(grad_(mu)n)_(nu):}\begin{equation*} K_{\alpha \beta}=-P_{\alpha}^{\mu} P_{\beta}^{\nu}\left(\boldsymbol{\nabla}_{\mu} \boldsymbol{n}\right)_{\nu} \tag{30.87} \end{equation*}(30.87)Kαβ=PαμPβν(μn)ν
(d) Show finally that, for vector field Y Y Y\boldsymbol{Y}Y that is also tangent to Σ Σ Sigma\SigmaΣ, we have
(30.88) Y X = D Y X + K ( X , Y ) n (30.88) Y X = D Y X + K ( X , Y ) n {:(30.88)grad_(Y)X=D_(Y)X+K(X","Y)n:}\begin{equation*} \nabla_{Y} X=D_{Y} X+K(X, Y) n \tag{30.88} \end{equation*}(30.88)YX=DYX+K(X,Y)n
Hint: The first term is clearly the part tangent to Σ Σ Sigma\SigmaΣ. To obtain the second term, consider the normal component of P μ α ( μ X ) β P μ α μ X β P^(mu)_(alpha)(grad_(mu)X)^(beta)P^{\mu}{ }_{\alpha}\left(\boldsymbol{\nabla}_{\mu} \boldsymbol{X}\right)^{\beta}Pμα(μX)β.
This formalism is useful in that we can use it to describe the Riemann tensor in terms of quantities projected into the hyperspace Σ Σ Sigma\SigmaΣ. These are the Gauss equation
P α μ P β ν P γ σ P δ ρ R μ ν σ ρ (30.89) = ( 3 ) R α β γ δ K α δ K β γ + K α γ K β δ , P α μ P β ν P γ σ P δ ρ R μ ν σ ρ (30.89) = ( 3 ) R α β γ δ K α δ K β γ + K α γ K β δ , {:[P_(alpha)^(mu)P_(beta)^(nu)P_(gamma)^(sigma)P_(delta)^(rho)R_(mu nu sigma rho)],[(30.89)=^((3))R_(alpha beta gamma delta)-K_(alpha delta)K_(beta gamma)+K_(alpha gamma)K_(beta delta)","]:}\begin{align*} & P_{\alpha}^{\mu} P_{\beta}^{\nu} P_{\gamma}^{\sigma} P_{\delta}^{\rho} R_{\mu \nu \sigma \rho} \\ & ={ }^{(3)} R_{\alpha \beta \gamma \delta}-K_{\alpha \delta} K_{\beta \gamma}+K_{\alpha \gamma} K_{\beta \delta}, \tag{30.89} \end{align*}PαμPβνPγσPδρRμνσρ(30.89)=(3)RαβγδKαδKβγ+KαγKβδ,
and the Codazzi equation
P α μ P β ν P γ σ n ρ R μ ν σ ρ = D β K α γ D α K β γ . P α μ P β ν P γ σ n ρ R μ ν σ ρ = D β K α γ D α K β γ . P_(alpha)^(mu)P_(beta)^(nu)P_(gamma)^(sigma)n^(rho)R_(mu nu sigma rho)=D_(beta)K_(alpha gamma)-D_(alpha)K_(beta gamma).P_{\alpha}^{\mu} P_{\beta}^{\nu} P_{\gamma}^{\sigma} n^{\rho} R_{\mu \nu \sigma \rho}=D_{\beta} K_{\alpha \gamma}-D_{\alpha} K_{\beta \gamma} .PαμPβνPγσnρRμνσρ=DβKαγDαKβγ.

31

31.1 Old notions of vectors and gradients
31.2 Vectors and vector fields 324 31.3 Linear slot machines again

31.4 Tensors again

1 1 ^(1){ }^{1}1 That is to say, we leave out the struc ture given by the metric, its vectorspace structure and also its topological structure. We shall gradually reintroduce these features in the following chapters.
Fig. 31.1 Left: some manifolds, smooth enough that the region around any point looks locally flat; right: some non-manifolds with points whose neighbourhood does not look smooth at any level of magnification.

A reintroduction to geometry

It is no matter what you teach them first, any more than what leg you shall put into your breeches first.
Samuel Johnson (1709-1784)
In this chapter, we once again meet the geometry of vectors and 1forms. Our goal is to more thoroughly define an approach that is free from the shackles of coordinates. This will hinge on the observation that each vector is equivalent to a derivative.
Although not essential for our purposes here, it is worth keeping the idea in mind that the arena in which we work in this chapter is a manifold. This is a space, often called M M M\mathcal{M}M, that is smooth: locally resembling the smoothness of R n R n R^(n)\mathbb{R}^{n}Rn, the usual Euclidean (i.e. flat) n n nnn-dimensional space. It is this smoothness that characterizes a manifold, since we leave out the rest of the rich structure 1 1 ^(1){ }^{1}1 of R n R n R^(n)\mathbb{R}^{n}Rn. In this smooth space, there are points called things like P , Q , A P , Q , A P,Q,A\mathcal{P}, \mathcal{Q}, \mathcal{A}P,Q,A and B B B\mathcal{B}B. We can cover patches of this space with coordinates, so that P P P\mathcal{P}P can be described by a set of n n nnn coordinates ( x 1 x n ) x 1 x n (x^(1)dotsx^(n))\left(x^{1} \ldots x^{n}\right)(x1xn), although this won't always be necessary. Curves are well-defined objects in our manifolds, parametrized by some quantity λ λ lambda\lambdaλ that varies monotonically along the curve. Differentiation, which measures changes along curves, is also a well-defined operation.

Example 31.1

It is useful to consider which spaces are smooth enough to qualify as a manifold. Examples of spaces that are manifolds include all of the n n nnn-dimensional Euclidean spaces called R n R n R^(n)\mathbb{R}^{n}Rn, which include R 1 R 1 R^(1)\mathbb{R}^{1}R1 (a line), R 2 R 2 R^(2)\mathbb{R}^{2}R2 (a plane) etc. (These clearly are locally identical to R n R n R^(n)\mathbb{R}^{n}Rn, since they are the spaces R n R n R^(n)\mathbb{R}^{n}Rn.) Other good examples which are smooth enough to look locally flat are (i) the one-dimensional circle, called S 1 S 1 S^(1)S^{1}S1; (ii) the two-dimensional surface of a sphere, called S 2 S 2 S^(2)S^{2}S2; and (iii) the two-dimensional surface of a torus. Some of these are shown on the left-hand side of Fig. 31.1. Perhaps more illuminating are spaces that aren't smooth enough to qualify as manifolds. Examples of these include (i) a line that juts out of a plane or (ii) a double cone; (iii) a line with a kink; (iv) a line that crosses itself. All of these structures have a point which even when blown up to a very large size, will never look locally flat. These latter examples are shown on the right-hand side of Fig. 31.1.
Our first task in this chapter will be to identify vectors and 1 -forms. These don't actually live in the manifold M M M\mathcal{M}M itself, but in related manifolds. Specifically, a vector defined at a point P P P\mathcal{P}P lives a manifold called
a tangent space, while 1-forms live in a dual space. For now, we will keep things as simple as possible and put these subtleties aside, retaining only the notion of points in a smooth manifold M M M\mathcal{M}M that, locally, looks like flat Euclidean space. So forget, temporarily, about the metric, and also the covariant derivatives and curvature tensors that the metric field can generate, as we look at how a smooth space, with a minimum of structure, can host a powerful geometry that will allow us many insights into the physics of relativity.

31.1 Old notions of vectors and gradients

We want to define vectors and describe surfaces in space. Let's review the old technology for doing this.

Example 31.2

The first time we meet vectors we are taught to picture them as a directed straight line linking two points. For example, the line from point A A A\mathcal{A}A to point B B B\mathcal{B}B (Fig. 31.2). We then write a vector in terms of coordinates as
(31.1) u = u α e α = u 0 e 0 + u 1 e 1 + u 2 e 2 + u 3 e 3 + (31.1) u = u α e α = u 0 e 0 + u 1 e 1 + u 2 e 2 + u 3 e 3 + {:(31.1)u=u^(alpha)e_(alpha)=u^(0)e_(0)+u^(1)e_(1)+u^(2)e_(2)+u^(3)e_(3)+dots:}\begin{equation*} \boldsymbol{u}=u^{\alpha} \boldsymbol{e}_{\alpha}=u^{0} e_{0}+u^{1} e_{1}+u^{2} e_{2}+u^{3} e_{3}+\ldots \tag{31.1} \end{equation*}(31.1)u=uαeα=u0e0+u1e1+u2e2+u3e3+
or u = u α e α u = u α e α u=u^(alpha)e_(alpha)\boldsymbol{u}=u^{\alpha} \boldsymbol{e}_{\alpha}u=uαeα, where the e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα make an appropriate basis, spanning the space in which we're working. Often we use an orthonormal basis such that e α e β = δ α β e α e β = δ α β e_(alpha)*e_(beta)=delta_(alpha beta)\boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta}=\delta_{\alpha \beta}eαeβ=δαβ.
If we are presented with a curved surface defined, in three spatial dimensions, by a function such as z = f ( x , y ) z = f ( x , y ) z=f(x,y)z=f(x, y)z=f(x,y), a particularly useful object is the gradient vector of the surface. We find this by rewriting f f fff as a function h ( x , y , z ) = 0 h ( x , y , z ) = 0 h(x,y,z)=0h(x, y, z)=0h(x,y,z)=0, whose gradient vector is
(31.2) h ( x , y , z ) = h x e x + h y e y + h z e z . (31.2) h ( x , y , z ) = h x e x + h y e y + h z e z . {:(31.2)grad h(x","y","z)=(del h)/(del x)e_(x)+(del h)/(del y)e_(y)+(del h)/(del z)e_(z).:}\begin{equation*} \boldsymbol{\nabla} h(x, y, z)=\frac{\partial h}{\partial x} \boldsymbol{e}_{x}+\frac{\partial h}{\partial y} \boldsymbol{e}_{y}+\frac{\partial h}{\partial z} \boldsymbol{e}_{z} . \tag{31.2} \end{equation*}(31.2)h(x,y,z)=hxex+hyey+hzez.
More generally we could write
(31.3) h = h x α e α . (31.3) h = h x α e α . {:(31.3)grad h=(del h)/(delx^(alpha))e_(alpha).:}\begin{equation*} \nabla h=\frac{\partial h}{\partial x^{\alpha}} e_{\alpha} . \tag{31.3} \end{equation*}(31.3)h=hxαeα.
This defines 2 2 ^(2){ }^{2}2 a vector pointing normal to the tangent plane of the surface. If we combine a vector v v v\boldsymbol{v}v and the gradient vector using the dot product, we obtain a useful object: the directional derivative of h h hhh, often denoted v h v h del_(v)h\partial_{\boldsymbol{v}} hvh, given by
(31.4) v h = v h = v α h x α (31.4) v h = v h = v α h x α {:(31.4)del_(v)h=v*grad h=v^(alpha)(del h)/(delx^(alpha)):}\begin{equation*} \partial_{\boldsymbol{v}} h=\boldsymbol{v} \cdot \boldsymbol{\nabla} h=v^{\alpha} \frac{\partial h}{\partial x^{\alpha}} \tag{31.4} \end{equation*}(31.4)vh=vh=vαhxα
which tells us the change in the function h h hhh along the direction of the vector v v v\boldsymbol{v}v. That is, the value of h h hhh at the tip of the vector v v v\boldsymbol{v}v, minus the value of h h hhh at the base of v v v\boldsymbol{v}v. The output of the directional derivative is a number.
For our purposes this old technology simply will not do. However, these ideas, when taken together with some new ones, allow us to tighten up the definitions and come up with a more useful geometry. For example, we will see how having 1 -forms allows us to abandon this rather forced (and non-covariant) notion of the gradient outputting a vector normal to a surface.
Fig. 31.2 A vector stretching between points A A A\mathcal{A}A and B B B\mathcal{B}B.
2 2 ^(2){ }^{2}2 Interpreting this as a vector immediately looks wrong from our tensor conventions of summing over one up and down index, since it appears that we have components h , α e α h , α e α h_(,alpha)e_(alpha)h_{, \alpha} \boldsymbol{e}_{\alpha}h,αeα.
3 3 ^(3){ }^{3}3 In words, input a value of λ λ lambda\lambdaλ and output a position P P P\mathcal{P}P on the curve.
Fig. 31.3 A vector as a derivative of
Fig. 31.4 A vector as a derivative in a coordinate system. We use the relationship between the point and the coordinates and also between the coordinates and the parametrization of the curve These are linked by the chain rule. 4 Or 4 Or ^(4)Or{ }^{4} \mathrm{Or}4Or, if you prefer, the tangent vector v v vvv is a derivative d / d λ d / d λ d//dlambda\mathrm{d} / \mathrm{d} \lambdad/dλ.

31.2 Vectors and vector fields

We often think of a vector as a directed line B A B A B-A\mathcal{B}-\mathcal{A}BA, joining two points A A A\mathcal{A}A and B B B\mathcal{B}B as in Fig. 31.2. Defining a vector in terms of two points is cumbersome. We would prefer to have the concept of a vector at a single point P P P\mathcal{P}P. Let's parametrize a straight line with a parameter λ λ lambda\lambdaλ by writing P ( λ ) = A + λ ( B A ) P ( λ ) = A + λ ( B A ) P(lambda)=A+lambda(B-A)\mathcal{P}(\lambda)=\mathcal{A}+\lambda(\mathcal{B}-\mathcal{A})P(λ)=A+λ(BA). This allows us to extract the vector as the difference between tip and base via
(31.5) d P ( λ ) d λ = B A (31.5) d P ( λ ) d λ = B A {:(31.5)(dP(lambda))/(dlambda)=B-A:}\begin{equation*} \frac{\mathrm{d} \mathcal{P}(\lambda)}{\mathrm{d} \lambda}=\mathcal{B}-\mathcal{A} \tag{31.5} \end{equation*}(31.5)dP(λ)dλ=BA
This idea of a vector as equivalent to a derivative is the key one in this chapter.
Advancing beyond straight lines, we can define a curve in terms of a another parametrized path 3 P ( λ ) 3 P ( λ ) ^(3)P(lambda){ }^{3} \mathcal{P}(\lambda)3P(λ), as we have in Fig. 31.3. The derivative formulation of vectors in the previous equation allows us to define a tangent vector to this curve via
(31.6) v = d P ( λ ) d λ (31.6) v = d P ( λ ) d λ {:(31.6)v=(dP(lambda))/(dlambda):}\begin{equation*} \boldsymbol{v}=\frac{\mathrm{d} \mathcal{P}(\lambda)}{\mathrm{d} \lambda} \tag{31.6} \end{equation*}(31.6)v=dP(λ)dλ
An interpretation of this expression, due to Elie Cartan, is that the vector represents the movement of the point P P P\mathcal{P}P with the derivative evaluating the difference between the point at the tip of the tangent vector, and the point at the base of this vector. Of course, we don't only have access to the parameter λ λ lambda\lambdaλ. We often describe curves in the manner shown in Fig. 31.4 using a coordinate system like x α x α x^(alpha)x^{\alpha}xα. Using these coordinates, the parametrized path is written as x α ( λ ) x α ( λ ) x^(alpha)(lambda)x^{\alpha}(\lambda)xα(λ), and we use the chain rule to say that our vector is expressed as
(31.7) v = d P d λ = d x α d λ P x α (31.7) v = d P d λ = d x α d λ P x α {:(31.7)v=(dP)/((d)lambda)=(dx^(alpha))/(dlambda)(delP)/(delx^(alpha)):}\begin{equation*} \boldsymbol{v}=\frac{\mathrm{d} \mathcal{P}}{\mathrm{~d} \lambda}=\frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \lambda} \frac{\partial \mathcal{P}}{\partial x^{\alpha}} \tag{31.7} \end{equation*}(31.7)v=dP dλ=dxαdλPxα
The next step is to realize that, instead of interpreting a vector as the movement of the point P P P\mathcal{P}P, we could strip off the point to make a more general vector operator along the curve
(31.8) v [ ] = d d λ = d x α d λ x α (31.8) v [ ] = d d λ = d x α d λ x α {:(31.8)v[]=(d)/((d)lambda)=(dx^(alpha))/(dlambda)(del)/(delx^(alpha)):}\begin{equation*} \boldsymbol{v}[]=\frac{\mathrm{d}}{\mathrm{~d} \lambda}=\frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \lambda} \frac{\partial}{\partial x^{\alpha}} \tag{31.8} \end{equation*}(31.8)v[]=d dλ=dxαdλxα
where the brackets in v [ v [ v[\boldsymbol{v}[v[ ] indicate that we need to provide a point on the curve as an input to the vector operator. In other words, the tangent vector v v v\boldsymbol{v}v at a point is identified with a derivative d / d λ d / d λ d//dlambda\mathrm{d} / \mathrm{d} \lambdad/dλ at that point. 4 4 ^(4){ }^{4}4 Therefore, we don't need to view the vector as an arrow representing the movement of a point, but rather as an object that is attached to a particular point on the curve. The vector will vary as you evaluate it at different points along the curve.
Finally, we return to our old notion of a vector, written as v = v α e α v = v α e α v=v^(alpha)e_(alpha)\boldsymbol{v}=v^{\alpha} \boldsymbol{e}_{\alpha}v=vαeα. Comparison with our new formulation of a vector in eqn 31.8 allows us to identify components and basis vectors as follows:
(31.9) d x α d λ = v α , x α = e α (31.9) d x α d λ = v α , x α = e α {:(31.9)(dx^(alpha))/(dlambda)=v^(alpha)","quad(del)/(delx^(alpha))=e_(alpha):}\begin{equation*} \frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \lambda}=v^{\alpha}, \quad \frac{\partial}{\partial x^{\alpha}}=e_{\alpha} \tag{31.9} \end{equation*}(31.9)dxαdλ=vα,xα=eα
where we note that e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα could be written as e α [ ] e α [ ] e_(alpha)[]\boldsymbol{e}_{\alpha}[]eα[], since this is the part in which we input points on the curve. 5 5 ^(5){ }^{5}5 Mixing this notation, allows us to write our vector as
(31.10) v [ ] d d λ = v α x α = v (31.10) v [ ] d d λ = v α x α = v {:(31.10)v[]-=(d)/((d)lambda)=v^(alpha)(del)/(delx^(alpha))=del_(v):}\begin{equation*} \boldsymbol{v}[] \equiv \frac{\mathrm{d}}{\mathrm{~d} \lambda}=v^{\alpha} \frac{\partial}{\partial x^{\alpha}}=\partial_{\boldsymbol{v}} \tag{31.10} \end{equation*}(31.10)v[]d dλ=vαxα=v
where we've introduced the directional derivative operator v v del_(v)\partial_{v}v in the last step.
We therefore have a new definition of a vector as identified with a directional derivative operator. This is sloganized as every vector can be identified with a derivative. We summarize our progress in the next box. A vector v v v\boldsymbol{v}v that is tangent to a curve P ( λ ) P ( λ ) P(lambda)\mathcal{P}(\lambda)P(λ) can be written as a derivative operator
(31.11) v [ ] = d d λ = d x α d λ x α = v α e α (31.11) v [ ] = d d λ = d x α d λ x α = v α e α {:(31.11)v[]=(d)/((d)lambda)=(dx^(alpha))/(dlambda)*(del)/(delx^(alpha))=v^(alpha)e_(alpha):}\begin{equation*} \boldsymbol{v}[]=\frac{\mathrm{d}}{\mathrm{~d} \lambda}=\frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \lambda} \cdot \frac{\partial}{\partial x^{\alpha}}=v^{\alpha} e_{\alpha} \tag{31.11} \end{equation*}(31.11)v[]=d dλ=dxαdλxα=vαeα
where λ λ lambda\lambdaλ parametrizes the curve. The vector's components are v α = v α = v^(alpha)=v^{\alpha}=vα= d x α d λ d x α d λ (dx^(alpha))/(dlambda)\frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \lambda}dxαdλ and the basis vectors are e α = x α e α = x α e_(alpha)=(del)/(delx^(alpha))\boldsymbol{e}_{\alpha}=\frac{\partial}{\partial x^{\alpha}}eα=xα.
The derivative in the definition of the vector could continue to be fed points like P P P\mathcal{P}P on the curve as it was in eqn 31.6. However, its real value is that we can input functions into this vector operator. We shall write the action of a vector field on a function 6 f ( x ) 6 f ( x ) ^(6)f(x){ }^{6} f(x)6f(x) as
v [ f ( x ) ] = d f ( x ) d λ = d x α d λ f ( x ) x α (31.12) = v α f ( x ) x α = v f ( x ) v [ f ( x ) ] = d f ( x ) d λ = d x α d λ f ( x ) x α (31.12) = v α f ( x ) x α = v f ( x ) {:[v[f(x)]=(df(x))/(dlambda)=(dx^(alpha))/(dlambda)*(del f(x))/(delx^(alpha))],[(31.12)=v^(alpha)(del f(x))/(delx^(alpha))=del_(v)f(x)]:}\begin{align*} \boldsymbol{v}[f(x)]=\frac{\mathrm{d} f(x)}{\mathrm{d} \lambda} & =\frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \lambda} \cdot \frac{\partial f(x)}{\partial x^{\alpha}} \\ & =v^{\alpha} \frac{\partial f(x)}{\partial x^{\alpha}}=\partial_{\boldsymbol{v}} f(x) \tag{31.12} \end{align*}v[f(x)]=df(x)dλ=dxαdλf(x)xα(31.12)=vαf(x)xα=vf(x)
This object evaluates the change in the function f ( x ) f ( x ) f(x)f(x)f(x) along the direction of the vector v v v\boldsymbol{v}v, that is, the directional derivative. 7 7 ^(7){ }^{7}7 So to recap:
Every vector is equivalent to a differential operator that inputs a function and outputs the directional derivative with respect to the vector. This idea that every vector is a derivative links the concepts of geometry and analysis, as shown in Fig. 31.5. This helps explain the power of these techniques in the study of general relativity.
We can show that tangent vectors defined via derivatives evaluated at a point P P P\mathcal{P}P, give a vector space. This relies on the notion that lots of curves can pass through a given point P P P\mathcal{P}P, allowing us to compare the different tangent vectors to each curve evaluated at P P P\mathcal{P}P (Fig. 31.6).
Example 31.3
A vector u u u\boldsymbol{u}u, tangent to the curve parametrized by μ μ mu\muμ, is written in coordinate space
(31.13) u = d d μ = d x α d μ x α (31.13) u = d d μ = d x α d μ x α {:(31.13)u=(d)/((d)mu)=(dx^(alpha))/(dmu)*(del)/(delx^(alpha)):}\begin{equation*} \boldsymbol{u}=\frac{\mathrm{d}}{\mathrm{~d} \mu}=\frac{\mathrm{d} x^{\alpha}}{\mathrm{d} \mu} \cdot \frac{\partial}{\partial x^{\alpha}} \tag{31.13} \end{equation*}(31.13)u=d dμ=dxαdμxα
Consider adding two different vectors at the same point P P P\mathcal{P}P (Fig. 31.6). In the derivative language, these vectors are tangent to two different curves at P P P\mathcal{P}P : a curve x ( λ ) x ( λ ) x(lambda)x(\lambda)x(λ) and a curve x ( μ ) x ( μ ) x(mu)x(\mu)x(μ). We write a linear combination, in terms of constants a a aaa and b b bbb, as
(31.14) a v + b u = a d d λ + b d d μ = ( a d x α d λ + b d x α d μ ) x α (31.14) a v + b u = a d d λ + b d d μ = a d x α d λ + b d x α d μ x α {:(31.14)av+bu=a((d))/((d)lambda)+b((d))/((d)mu)=(a((d)x^(alpha))/(dlambda)+b((d)x^(alpha))/(dmu))(del)/(delx^(alpha)):}\begin{equation*} a \boldsymbol{v}+b \boldsymbol{u}=a \frac{\mathrm{~d}}{\mathrm{~d} \lambda}+b \frac{\mathrm{~d}}{\mathrm{~d} \mu}=\left(a \frac{\mathrm{~d} x^{\alpha}}{\mathrm{d} \lambda}+b \frac{\mathrm{~d} x^{\alpha}}{\mathrm{d} \mu}\right) \frac{\partial}{\partial x^{\alpha}} \tag{31.14} \end{equation*}(31.14)av+bu=a d dλ+b d dμ=(a dxαdλ+b dxαdμ)xα
5 5 ^(5){ }^{5}5 In other words, the components tell us how the coordinates x α x α x^(alpha)x^{\alpha}xα change with the curve's parameter λ λ lambda\lambdaλ; the basis vectors represent the rate of change with respect to our choice of coordinates.
6 6 ^(6){ }^{6}6 We continue to use square brackets for this, and round brackets for the vector's slot that inputs a 1 -form.
7 7 ^(7){ }^{7}7 It might sometimes help to continue to think of this as the value of the function at the tip of the vector, minus the value at the base of the vector.
Fig. 31.5 Links between geometry and analysis.
Fig. 31.6 Two of the many curves passing through point P P P\mathcal{P}P parametrized by λ λ lambda\lambdaλ and μ μ mu\muμ, respectively. Their tangents at P P P\mathcal{P}P are shown.
8 8 ^(8){ }^{8}8 Roughly speaking, a vector space is characterized by linear combinations of characterized by linear combinations of
vectors and has the property of closure vectors and has the property of closure
(so that any linear combination of vectors belonging to the vector space can only produce a vector which is itself a member of the vector space). The defining features of a vector space include the existence of a zero vector and the existence of an inverse of any nonzero vector, and also properties such as associativity, distributivity and comas associativity, distributivity and commutativity. There must be a basis of linearly independent vectors that spans
the space, the number of which are the space, the number of which are
equal to the number of dimensions of the space.
9 9 ^(9){ }^{9}9 We write this d / d λ | P d / d λ P d//d lambda|_(P)d /\left.d \lambda\right|_{\mathcal{P}}d/dλ|P to remove any ambiguity about which point we mean Note that we don't write this d P / d λ d P / d λ dP//dlambda\mathrm{d} \mathcal{P} / \mathrm{d} \lambdadP/dλ since we don't any longer need to input the point P P P\mathcal{P}P into the derivative and generally, we won't. The vector will most often operate on a function to deliver the function's directional derivative along the vector.
10 10 ^(10){ }^{10}10 These are important in physics. For example, we shall see in Chapter 39 how the congruence represents the streamlines of a fluid.
Fig. 31.7 A congruence of curves fills the space but the curves never intersect. Their tangent vectors form a vector field.
11 11 ^(11){ }^{11}11 In the following chapters, we will see that there is another derivative that can also be used to compare vector fields at different points: this is the so-called Lie derivative.
where we've expanded both vectors in terms of our choice of coordinates x α x α x^(alpha)x^{\alpha}xα in the second step, so that they're referred to the same set of basis vectors e α = / x α e α = / x α e_(alpha)=del//delx^(alpha)\boldsymbol{e}_{\alpha}=\partial / \partial x^{\alpha}eα=/xα. The previous equation defines a new vector w w w\boldsymbol{w}w, with components ( a d x α d λ + b d x α d μ ) a d x α d λ + b d x α d μ (a((d)x^(alpha))/(dlambda)+b((d)x^(alpha))/(dmu))\left(a \frac{\mathrm{~d} x^{\alpha}}{\mathrm{d} \lambda}+b \frac{\mathrm{~d} x^{\alpha}}{\mathrm{d} \mu}\right)(a dxαdλ+b dxαdμ). This new vector must be tangent to some other curve through P P P\mathcal{P}P with, say, a parameter ϕ ϕ phi\phiϕ, so we write
We conclude that
d d ϕ = a d d λ + b d d μ d d ϕ = a d d λ + b d d μ (d)/((d)phi)=a((d))/((d)lambda)+b((d))/((d)mu)\frac{\mathrm{d}}{\mathrm{~d} \phi}=a \frac{\mathrm{~d}}{\mathrm{~d} \lambda}+b \frac{\mathrm{~d}}{\mathrm{~d} \mu}d dϕ=a d dλ+b d dμ
or w = a v + b u w = a v + b u w=av+bu\boldsymbol{w}=a \boldsymbol{v}+b \boldsymbol{u}w=av+bu. This means that the directions dives for indeed space 8 8 ^(8)^{8}8 at P P P\mathcal{P}P. Why? It is because linear combinations of a set of vectors ( d / d λ d / d λ d//dlambda\mathrm{d} / \mathrm{d} \lambdad/dλ and d / d μ d / d μ d//dmu\mathrm{d} / \mathrm{d} \mud/dμ here) at P P P\mathcal{P}P can express the directional derivatives of other curves at this point.
To summarize, our new formulation of vectors as derivative operators has been based on evaluating tangent vectors along a curve at a point such as P P P\mathcal{P}P. The vector v = d / d λ | P v = d / d λ P v=d//dlambda|_(P)v=\mathrm{d} /\left.\mathrm{d} \lambda\right|_{\mathcal{P}}v=d/dλ|P is tangent to the curve parametrized by λ λ lambda\lambdaλ at point 9 P 9 P ^(9)P{ }^{9} \mathcal{P}9P. Many other curves pass through point P P P\mathcal{P}P and we have seen how the vectors defined at this point make a vector space. We can express an arbitrary vector at P P P\mathcal{P}P as a superposition of other vectors at P P P\mathcal{P}P.
Now let's imagine there being a large family of curves that fill the space, but don't intersect, as shown in Fig. 31.7. Each curve is parametrized by parameter λ λ lambda\lambdaλ. (However, a point such as λ = 2 λ = 2 lambda=2\lambda=2λ=2 should be expected to label points different distances along each curve.) Such a set of space-filling curves that never intersect each other is called a congruence. 10 10 ^(10){ }^{10}10 The existence of a congruence means that at any point in space there is a curve with a unique derivative. Since derivatives are equivalent to tangent vectors, this gives us the notion of a vector field: at any point in a space we can identify a unique vector. At point Q Q Q\mathcal{Q}Q, for example, we call that vector d / d λ | Q d / d λ Q d//d lambda|_(Q)d /\left.d \lambda\right|_{\mathcal{Q}}d/dλ|Q. It is the tangent to that curve from the congruence that exists at point Q Q Q\mathcal{Q}Q. Put the other way around, we can pick a point and the vector field will input this point and output a vector. The vector-field machine works out what the vector is by computing the tangent to that curve found in the congruence at that particular point.
In this formulation, it only makes sense to compare different vectors defined at a single point, such as Q Q Q\mathcal{Q}Q. There is no obvious way to compare vectors defined at another point P P P\mathcal{P}P with those at Q Q Q\mathcal{Q}Q. In order to do this, we would need a way of moving vectors around, such that they can be compared at a single point. The method of moving vectors would require the extra information on how vectors change as they move throughout the space, in order to make the comparison a meaningful one. In the previous chapters, we saw that this requires the notion of the connection that led to the covariant derivative. The connection allowed us to formulate a notion of parallelism: how to move a vector through space such that the only changes would be due to how that space varied. 11 11 ^(11){ }^{11}11 For now, we will continue to compare objects defined at only a single point.

31.3 Linear slot machines again

Next, we shall expand the menu of objects to which we have access by considering 1 -forms. These are members of a family of objects called 'differential forms' or simply 'forms' that will occupy us for the remainder of this part of the book. Forms were proposed (or discovered) by Elié Cartan in around 1900. While vectors have become an essential tool whose properties are taught to all undergraduate physicists, forms do not currently share this status. This is arguably a mistake, since they are no more complicated than vectors, and allow access to a far more simple and clear path to understanding geometry than is often presented.
We have, from our progress in this chapter, the notion of vector fields as derivatives that act on functions at a particular point in space. Quite separately from this, each vector has another possible input: vectors have a slot v ( ) v ( ) v()\boldsymbol{v}()v() that accepts a 1 -form and returns a scalar. Calling a typical 1-form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~, we have v ( σ ~ ) σ ~ , v = v ( σ ~ ) σ ~ , v = v( tilde(sigma))-=(: tilde(sigma),v:)=\boldsymbol{v}(\tilde{\boldsymbol{\sigma}}) \equiv\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{v}\rangle=v(σ~)σ~,v= (number). 12 12 ^(12){ }^{12}12
Vectors, 1 -forms and numbers live, in a sense, in different spaces. We mentioned how vectors live in a tangent space and, by the same token, 1 -forms live in a dual space. In addition, scalar numbers live on the real line R 1 R 1 R^(1)\mathbb{R}^{1}R1, giving a situation shown in Fig. 31.8. The important thing about the inner product summarized in the last equation is that it allows us to map between these spaces.
A 1-form maps a vector onto a number; a vector maps a 1 -form onto a number.
This mapping onto a number (i.e. a point on the real line) is what we mean when we say that a vector is dual to a 1 -form. An important property of the mapping/inner product operation is linearity, which is to say
(31.18) ( a σ ~ + b ρ ~ ) , v = a σ ~ , v + b ρ ~ , v (31.18) ( a σ ~ + b ρ ~ ) , v = a σ ~ , v + b ρ ~ , v {:(31.18)(:(a tilde(sigma)+b tilde(rho))","v:)=a(: tilde(sigma)","v:)+b(: tilde(rho)","v:):}\begin{equation*} \langle(a \tilde{\boldsymbol{\sigma}}+b \tilde{\boldsymbol{\rho}}), \boldsymbol{v}\rangle=a\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{v}\rangle+b\langle\tilde{\boldsymbol{\rho}}, \boldsymbol{v}\rangle \tag{31.18} \end{equation*}(31.18)(aσ~+bρ~),v=aσ~,v+bρ~,v
and similarly, σ ~ , ( n v + m u ) = n σ ~ , v + m σ ~ , u σ ~ , ( n v + m u ) = n σ ~ , v + m σ ~ , u (: tilde(sigma),(nv+mu):)=n(: tilde(sigma),v:)+m(: tilde(sigma),u:)\langle\tilde{\boldsymbol{\sigma}},(n \boldsymbol{v}+m \boldsymbol{u})\rangle=n\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{v}\rangle+m\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{u}\rangleσ~,(nv+mu)=nσ~,v+mσ~,u.
We saw in Chapter 4 that while a vector can be thought of as a directed arrow, a 1 -form can the thought of an infinite set of equally spaced surfaces (Fig. 31.9). In this picture, the inner product of a vector and a 1 -form outputs a number that corresponds to the number of surfaces that the vector pierces.

Example 31.4

Consider a 1 -form A ~ A ~ tilde(A)\tilde{\boldsymbol{A}}A~. Working in Euclidean space we find the number of surfaces pierced by a unit vector in the x x xxx-direction is A ~ , e 1 = A 1 A ~ , e 1 = A 1 (:( tilde(A)),e_(1):)=A_(1)\left\langle\tilde{\boldsymbol{A}}, \boldsymbol{e}_{1}\right\rangle=A_{1}A~,e1=A1. This implies that the spacing of the surfaces along this direction is 1 / ( A 1 ) 1 / A 1 1//(A_(1))1 /\left(A_{1}\right)1/(A1). Multiplying the form by some factor F F FFF causes the density of surfaces to be increased by the factor F F FFF.
Just as a vector can be written in terms of components and basis vectors, a 1 -form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ is written in terms of components σ α σ α sigma_(alpha)\sigma_{\alpha}σα and basis 1 -forms ω α ω α omega^(alpha)\boldsymbol{\omega}^{\alpha}ωα as σ ~ = σ α ω α σ ~ = σ α ω α tilde(sigma)=sigma_(alpha)omega^(alpha)\tilde{\boldsymbol{\sigma}}=\sigma_{\alpha} \boldsymbol{\omega}^{\alpha}σ~=σαωα. To allow the mapping between vectors, 1 -forms and
12 12 ^(12){ }^{12}12 We shall maintain the symmetry of the operation, so that the 1 -form has a slot in which we can insert a vector σ ~ ( v ) = σ ~ ( v ) = tilde(sigma)(v)=\tilde{\boldsymbol{\sigma}}(\boldsymbol{v})=σ~(v)= (number), where, in all cases we shall examine, v ( σ ~ ) = σ ~ ( v ) v ( σ ~ ) = σ ~ ( v ) v( tilde(sigma))= tilde(sigma)(v)\boldsymbol{v}(\tilde{\boldsymbol{\sigma}})=\tilde{\boldsymbol{\sigma}}(\boldsymbol{v})v(σ~)=σ~(v). This is just how we define the inner product is just how we define the inner product between a vector and a 1-form. This is sometimes denoted by a dot, and when the dot would be confusing (as it usually denotes the scalar product of two vectors), more often by angle brackets σ ~ v σ ~ , v σ ~ v σ ~ , v tilde(sigma)*v-=(: tilde(sigma),v:)\tilde{\boldsymbol{\sigma}} \cdot \boldsymbol{v} \equiv\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{v}\rangleσ~vσ~,v. In summary,
σ ~ v σ ~ , v = v ( σ ~ ) = σ ~ ( v ) σ ~ v σ ~ , v = v ( σ ~ ) = σ ~ ( v ) tilde(sigma)*v-=(: tilde(sigma),v:)=v( tilde(sigma))= tilde(sigma)(v)\tilde{\boldsymbol{\sigma}} \cdot \boldsymbol{v} \equiv\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{v}\rangle=\boldsymbol{v}(\tilde{\boldsymbol{\sigma}})=\tilde{\boldsymbol{\sigma}}(\boldsymbol{v})σ~vσ~,v=v(σ~)=σ~(v)
= (number) =  (number)  =" (number) "=\text { (number) }= (number) 
(31.17)
Fig. 31.8 Vectors live in a tangent space, 1 -forms in a dual space and numbers along the real line R 1 R 1 R^(1)\mathbb{R}^{1}R1. A vector maps a 1 -form onto a number; a 1 -form maps a vector on to a number
Fig. 31.9 The 1-form A ~ A ~ tilde(A)\tilde{\boldsymbol{A}}A~ represented as a set of repeating surfaces. The inner product A ^ , X A ^ , X (: hat(A),X:)\langle\hat{\boldsymbol{A}}, \boldsymbol{X}\rangleA^,X tells us how many surfaces are pierced by vector X X X\boldsymbol{X}X.
13 13 ^(13){ }^{13}13 Recall that the differential d f d f df\mathrm{d} fdf of a function f ( x , y ) f ( x , y ) f(x,y)f(x, y)f(x,y) is written as
d f = ( f x ) y d x + ( f y ) x d y d f = f x y d x + f y x d y df=((del f)/(del x))_(y)dx+((del f)/(del y))_(x)dy\begin{equation*} \mathrm{d} f=\left(\frac{\partial f}{\partial x}\right)_{y} \mathrm{~d} x+\left(\frac{\partial f}{\partial y}\right)_{x} \mathrm{~d} y \tag{} \end{equation*}df=(fx)y dx+(fy)x dy
We use a bold d d d\boldsymbol{d}d here to denote the
differential operation. The reason for this will be explained in a few Chapters' time.
14 14 ^(14){ }^{14}14 As a result of this manipulation we are able to effectively retire the vector's square bracket slot and rely on the combination of the vector and 1 -form to make numbers via tensor operations.
numbers, we define an inner product between basis 1 -forms and basis vectors as
(31.19) ω β , e α = δ β α . (31.19) ω β , e α = δ β α . {:(31.19)(:omega^(beta),e_(alpha):)=delta^(beta)_(alpha).:}\begin{equation*} \left\langle\boldsymbol{\omega}^{\beta}, \boldsymbol{e}_{\alpha}\right\rangle=\delta^{\beta}{ }_{\alpha} . \tag{31.19} \end{equation*}(31.19)ωβ,eα=δβα.
In the new philosophy of having a vector correspond to a derivative, we have basis vectors e μ = x μ e μ = x μ e_(mu)=(del)/(delx^(mu))\boldsymbol{e}_{\mu}=\frac{\partial}{\partial x^{\mu}}eμ=xμ, where x μ x μ x^(mu)x^{\mu}xμ are a set of coordinates. We might also ask ourselves whether the 1 -form corresponds to another operation? Of course it does: it is the operation that is dual to the operation of taking a derivative. Specifically, each basis 1-form corresponds to a differential 13 13 ^(13){ }^{13}13
(31.21) ω α = d x α . (31.21) ω α = d x α . {:(31.21)omega^(alpha)=dx^(alpha).:}\begin{equation*} \boldsymbol{\omega}^{\alpha}=\boldsymbol{d} x^{\alpha} . \tag{31.21} \end{equation*}(31.21)ωα=dxα.
This works in our coordinate basis since
(31.22) ω μ , e ν = d x μ , x ν = x μ x ν = δ μ ν . (31.22) ω μ , e ν = d x μ , x ν = x μ x ν = δ μ ν . {:(31.22)(:omega^(mu),e_(nu):)=(:dx^(mu),(del)/(delx^(nu)):)=(delx^(mu))/(delx^(nu))=delta^(mu)_(nu).:}\begin{equation*} \left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{e}_{\nu}\right\rangle=\left\langle\boldsymbol{d} x^{\mu}, \frac{\partial}{\partial x^{\nu}}\right\rangle=\frac{\partial x^{\mu}}{\partial x^{\nu}}=\delta^{\mu}{ }_{\nu} . \tag{31.22} \end{equation*}(31.22)ωμ,eν=dxμ,xν=xμxν=δμν.
The simplest example of 1-form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ is the differential of a function f ( x ) f ( x ) f(x)f(x)f(x), which we write
σ ~ = d f = f x α d x α = f x α ω α . σ ~ = d f = f x α d x α = f x α ω α . tilde(sigma)=df=(del f)/(delx^(alpha))dx^(alpha)=(del f)/(delx^(alpha))omega^(alpha).\tilde{\boldsymbol{\sigma}}=\boldsymbol{d} f=\frac{\partial f}{\partial x^{\alpha}} \boldsymbol{d} x^{\alpha}=\frac{\partial f}{\partial x^{\alpha}} \boldsymbol{\omega}^{\alpha} .σ~=df=fxαdxα=fxαωα.
The identification of the 1 -form as a differential allows us to reconcile the two sets of slots we have given the vector u u u\boldsymbol{u}u : square brackets that accept a function and round ones that accept a 1 -form. We have that to operate on a function f ( x ) f ( x ) f(x)f(x)f(x) with the vector u [ ] u [ ] u[]\boldsymbol{u}[]u[], to make u [ f ] u [ f ] u[f]\boldsymbol{u}[f]u[f], we can equivalently take an inner product between the vector u u u\boldsymbol{u}u and the 1 -form d f d f df\boldsymbol{d} fdf, written as u ( d f ) u ( d f ) u(df)\boldsymbol{u}(\boldsymbol{d} f)u(df) or, more helpfully d f , u d f , u (:df,u:)\langle\boldsymbol{d} f, \boldsymbol{u}\rangledf,u. This is to say that we write 14 14 ^(14){ }^{14}14
u [ f ] d f , u = f x α ω α , u β e β = u β f x α ω α , e β (31.23) = u β f x α δ α β = u α f x α . u [ f ] d f , u = f x α ω α , u β e β = u β f x α ω α , e β (31.23) = u β f x α δ α β = u α f x α . {:[u[f]-=(:df","u:)=(:(del f)/(delx^(alpha))omega^(alpha),u^(beta)e_(beta):)],[=u^(beta)(del f)/(delx^(alpha))(:omega^(alpha),e_(beta):)],[(31.23)=u^(beta)(del f)/(delx^(alpha))delta^(alpha)_(beta)=u^(alpha)(del f)/(delx^(alpha)).]:}\begin{align*} \boldsymbol{u}[f] \equiv\langle\boldsymbol{d} f, \boldsymbol{u}\rangle & =\left\langle\frac{\partial f}{\partial x^{\alpha}} \boldsymbol{\omega}^{\alpha}, u^{\beta} \boldsymbol{e}_{\beta}\right\rangle \\ & =u^{\beta} \frac{\partial f}{\partial x^{\alpha}}\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\right\rangle \\ & =u^{\beta} \frac{\partial f}{\partial x^{\alpha}} \delta^{\alpha}{ }_{\beta}=u^{\alpha} \frac{\partial f}{\partial x^{\alpha}} . \tag{31.23} \end{align*}u[f]df,u=fxαωα,uβeβ=uβfxαωα,eβ(31.23)=uβfxαδαβ=uαfxα.
We conclude that the inner product gives the same directional derivative we had before (telling us how the function changes along the vector u u u\boldsymbol{u}u ) and so, in terms of the directional derivative operator, we write
(31.24) u [ f ] = u f = d f , u (31.24) u [ f ] = u f = d f , u {:(31.24)u[f]=del_(u)f=(:df","u:):}\begin{equation*} \boldsymbol{u}[f]=\partial_{\boldsymbol{u}} f=\langle\boldsymbol{d} f, \boldsymbol{u}\rangle \tag{31.24} \end{equation*}(31.24)u[f]=uf=df,u
Finally, we note that we have been dealing with vectors and 1 -forms defined at the same point in space P P P\mathcal{P}P. Just as we define a vector field we also define a 1 -form field. That is to say, at every point in space we have access to a unique 1 -form. This can be combined with the vector we obtain at this point from a vector field to make a number. The numbers evaluated at each point in space themselves form a scalar field.

31.4 Tensors again

Tensors can be regarded as generalized slot machines. 15 15 ^(15){ }^{15}15 They are machines in which we insert combinations of vectors and 1-forms in order to return numbers. Generally, a valence ( n , m ) ( n , m ) (n,m)(n, m)(n,m) tensor has slots for the insertion of n n nnn lots of 1 -forms and m m mmm lots of vectors, in order to return a number. We have the defining rule for a ( n , m ) ( n , m ) (n,m)(n, m)(n,m) tensor that
(31.25) T ( σ ~ 1 , , σ ~ n , v 1 , , v m ) = ( number ) (31.25) T σ ~ 1 , , σ ~ n , v 1 , , v m = (  number  ) {:(31.25)T( tilde(sigma)^(1),dots, tilde(sigma)^(n),v^(1),dots,v^(m))=(" number "):}\begin{equation*} \boldsymbol{T}\left(\tilde{\boldsymbol{\sigma}}^{1}, \ldots, \tilde{\boldsymbol{\sigma}}^{n}, \boldsymbol{v}^{1}, \ldots, \boldsymbol{v}^{m}\right)=(\text { number }) \tag{31.25} \end{equation*}(31.25)T(σ~1,,σ~n,v1,,vm)=( number )
Tensors can be built using a tensor product. The tensor product 16 16 ^(16){ }^{16}16 ox\otimes of two vectors gives us a ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor
(31.26) T ( , ) = u v = u ( ) v ( ) (31.26) T ( , ) = u v = u ( ) v ( ) {:(31.26)T(",")=u ox v=u()ox v():}\begin{equation*} \boldsymbol{T}(,)=\boldsymbol{u} \otimes \boldsymbol{v}=\boldsymbol{u}() \otimes \boldsymbol{v}() \tag{31.26} \end{equation*}(31.26)T(,)=uv=u()v()
This object has two slots into which 1-forms can be inserted. To find the components of the tensor (which are a set of numbers) insert basis 1 -forms into each of the slots.

Example 31.5

How do we extract the components of the (2,0) tensor T = u v T = u v T=u ox v\boldsymbol{T}=\boldsymbol{u} \otimes \boldsymbol{v}T=uv ? We just insert the basis 1 -forms into the slots. Here's how it goes.
T μ ν = T ( ω α , ω β ) = u ( ω α ) v ( ω β ) = ω α , u ω β , v (31.27) = u α v β . T μ ν = T ω α , ω β = u ω α v ω β = ω α , u ω β , v (31.27) = u α v β . {:[T^(mu nu)=T(omega^(alpha),omega^(beta))=u(omega^(alpha))ox v(omega^(beta))],[=(:omega^(alpha),u:)(:omega^(beta),v:)],[(31.27)=u^(alpha)v^(beta).]:}\begin{align*} T^{\mu \nu}=\boldsymbol{T}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{\omega}^{\beta}\right) & =\boldsymbol{u}\left(\boldsymbol{\omega}^{\alpha}\right) \otimes \boldsymbol{v}\left(\boldsymbol{\omega}^{\beta}\right) \\ & =\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{u}\right\rangle\left\langle\boldsymbol{\omega}^{\beta}, \boldsymbol{v}\right\rangle \\ & =u^{\alpha} v^{\beta} . \tag{31.27} \end{align*}Tμν=T(ωα,ωβ)=u(ωα)v(ωβ)=ωα,uωβ,v(31.27)=uαvβ.
We conclude T α β = u α v β T α β = u α v β T^(alpha beta)=u^(alpha)v^(beta)T^{\alpha \beta}=u^{\alpha} v^{\beta}Tαβ=uαvβ. We can reconstruct the tensor from the components using
(31.28) T = u α v β e α e β (31.28) T = u α v β e α e β {:(31.28)T=u^(alpha)v^(beta)e_(alpha)oxe_(beta):}\begin{equation*} \boldsymbol{T}=u^{\alpha} v^{\beta} \boldsymbol{e}_{\alpha} \otimes \boldsymbol{e}_{\beta} \tag{31.28} \end{equation*}(31.28)T=uαvβeαeβ
which is a shorthand for a version written with the slots:
(31.29) T ( , ) = u α v β e α ( ) e β ( ) (31.29) T ( , ) = u α v β e α ( ) e β ( ) {:(31.29)T(",")=u^(alpha)v^(beta)e_(alpha)()oxe_(beta)():}\begin{equation*} \boldsymbol{T}(,)=u^{\alpha} v^{\beta} \boldsymbol{e}_{\alpha}() \otimes \boldsymbol{e}_{\beta}() \tag{31.29} \end{equation*}(31.29)T(,)=uαvβeα()eβ()
We can use the tensor product to build objects out of 1-forms too.

Example 31.6

Consider a set of basis 1 -forms d x μ = ω μ d x μ = ω μ dx^(mu)=omega^(mu)\boldsymbol{d} x^{\mu}=\boldsymbol{\omega}^{\mu}dxμ=ωμ. The metric is a written as a ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor
(31.30) g = g μ ν d x μ d x ν = g μ ν ω μ ω ν (31.30) g = g μ ν d x μ d x ν = g μ ν ω μ ω ν {:(31.30)g=g_(mu nu)dx^(mu)ox dx^(nu)=g_(mu nu)omega^(mu)oxomega^(nu):}\begin{equation*} \boldsymbol{g}=g_{\mu \nu} \boldsymbol{d} x^{\mu} \otimes \boldsymbol{d} x^{\nu}=g_{\mu \nu} \boldsymbol{\omega}^{\mu} \otimes \boldsymbol{\omega}^{\nu} \tag{31.30} \end{equation*}(31.30)g=gμνdxμdxν=gμνωμων
This is sometimes called d s 2 d s 2 ds^(2)\boldsymbol{d} \boldsymbol{s}^{2}ds2. With this the metric becomes a two-slot, ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) object called g ( g ( g(\boldsymbol{g}(g(, ) , i n t o w h i c h w e i n s e r t v e c t o r s t o r e t u r n a n u m b e r . ) , i n t o w h i c h w e i n s e r t v e c t o r s t o r e t u r n a n u m b e r . ),intowhichweinsertvectorstoreturnanumber.) , into which we insert vectors to return a number.),intowhichweinsertvectorstoreturnanumber.
An arbitrary tensor ( n , m ) ( n , m ) (n,m)(n, m)(n,m) tensor can be built from n n nnn vectors and m m mmm 1-forms
(31.31) T = u 1 u n σ ~ 1 σ ~ m (31.31) T = u 1 u n σ ~ 1 σ ~ m {:(31.31)T=u^(1)ox dots oxu^(n)ox tilde(sigma)^(1)ox dots ox tilde(sigma)^(m):}\begin{equation*} \boldsymbol{T}=\boldsymbol{u}^{1} \otimes \ldots \otimes \boldsymbol{u}^{n} \otimes \tilde{\boldsymbol{\sigma}}^{1} \otimes \ldots \otimes \tilde{\boldsymbol{\sigma}}^{m} \tag{31.31} \end{equation*}(31.31)T=u1unσ~1σ~m
15 15 ^(15){ }^{15}15 Tensor calculus was developed mainly by Gregorio Ricci-Curbastro (1853-1925). His most famous work (1853-1925). His most famous work
on the calculus of tensors was coon the calculus of tensors was co-
authored with his former student, authored with his former student,
Tullio Levi-Civita (1873-1941), and signed Gregorio Ricci. (Curiously, all of Ricci's other works feature his full name, Ricci-Curbastro.) This work Méthods de calcul différential absolu et leurs applications from 1901 greatly simplified the presentation of Riemannian geometry and gave us the version of geometry that Einstein used as the basis for his general relativity 16 16 ^(16){ }^{16}16 Recall that ox\otimes simply tells us to retain the order of the slots.
17 17 ^(17){ }^{17}17 The tensor product sign tells us to maintain the order of vectors parts and, separately, the 1 -form parts. However, it doesn't matter if you list the basis vectors first or the basis 1 -forms first as long as you know which is which.
18 18 ^(18){ }^{18}18 Our presentation here follows Misner, Thorne and Wheeler, which can be consulted for further details.
To find the tensor's components by inserting n n nnn basis 1 -forms and m m mmm basis vectors:
(31.32) T ν 1 ν m μ 1 μ n = u 1 ( ω μ 1 ) u n ( ω μ n ) σ ~ 1 ( e ν 1 ) σ ~ m ( e ν m ) (31.32) T ν 1 ν m μ 1 μ n = u 1 ω μ 1 u n ω μ n σ ~ 1 e ν 1 σ ~ m e ν m {:(31.32)T_(nu_(1)dotsnu_(m))^(mu_(1)dotsmu_(n))=u^(1)(omega^(mu_(1)))dotsu^(n)(omega^(mu_(n))) tilde(sigma)^(1)(e_(nu_(1)))dots tilde(sigma)^(m)(e_(nu_(m))):}\begin{equation*} T_{\nu_{1} \ldots \nu_{m}}^{\mu_{1} \ldots \mu_{n}}=\boldsymbol{u}^{1}\left(\boldsymbol{\omega}^{\mu_{1}}\right) \ldots \boldsymbol{u}^{n}\left(\boldsymbol{\omega}^{\mu_{n}}\right) \tilde{\boldsymbol{\sigma}}^{1}\left(\boldsymbol{e}_{\nu_{1}}\right) \ldots \tilde{\boldsymbol{\sigma}}^{m}\left(\boldsymbol{e}_{\nu_{m}}\right) \tag{31.32} \end{equation*}(31.32)Tν1νmμ1μn=u1(ωμ1)un(ωμn)σ~1(eν1)σ~m(eνm)
If we know the components, we can recreate the tensor by expanding
(31.33) T = T ν 1 ν m μ 1 μ n ( e μ 1 e μ n ω ν 1 ω ν m ) . (31.33) T = T ν 1 ν m μ 1 μ n e μ 1 e μ n ω ν 1 ω ν m . {:(31.33)T=T_(nu_(1)dotsnu_(m))^(mu_(1)dotsmu_(n))(e_(mu_(1))ox dots oxe_(mu_(n))oxomega^(nu_(1))ox dots oxomega^(nu_(m))).:}\begin{equation*} \boldsymbol{T}=T_{\nu_{1} \ldots \nu_{m}}^{\mu_{1} \ldots \mu_{n}}\left(\boldsymbol{e}_{\mu_{1}} \otimes \ldots \otimes \boldsymbol{e}_{\mu_{n}} \otimes \boldsymbol{\omega}^{\nu_{1}} \otimes \ldots \otimes \boldsymbol{\omega}^{\nu_{m}}\right) . \tag{31.33} \end{equation*}(31.33)T=Tν1νmμ1μn(eμ1eμnων1ωνm).
We will frequently encounter tensors built using tensor products from a selection of vectors and 1-forms. One useful rule is that, although you can't alter the order of tensors or vectors amongst themselves, the 1 -form parts and vector parts commute. 17 17 ^(17){ }^{17}17
It should be no surprise that in the same way that we defined vector fields and 1-form fields, we can also define tensor fields. That is, at each point in space, we have a unique tensor that can be combined with vectors and 1 -forms found at that same point to make a number.

31.5 Examples of tensor operations

Once we have our tensors we can apply a range of tools to manipulate them. Some of the common operations are described in the example below. 18 18 ^(18){ }^{18}18
Example 31.7
Consider, for example, a tensor P P P\boldsymbol{P}P of rank ( 0 , 3 ) ( 0 , 3 ) (0,3)(0,3)(0,3), which we can write
(31.34) P ( , , ) = P α β γ ω α ω β ω γ (31.34) P ( , , ) = P α β γ ω α ω β ω γ {:(31.34)P(","",")=P_(alpha beta gamma)omega^(alpha)oxomega^(beta)oxomega^(gamma):}\begin{equation*} \boldsymbol{P}(,,)=P_{\alpha \beta \gamma} \boldsymbol{\omega}^{\alpha} \otimes \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{\omega}^{\gamma} \tag{31.34} \end{equation*}(31.34)P(,,)=Pαβγωαωβωγ
In words: P P P\boldsymbol{P}P has three slots which all take vectors. If we fill the slots with vectors we output a scalar given by
P ( u , v , w ) = P α β γ ω α ( u ) ω β ( v ) ω γ ( w ) (31.35) = P α β γ u α v β w γ P ( u , v , w ) = P α β γ ω α ( u ) ω β ( v ) ω γ ( w ) (31.35) = P α β γ u α v β w γ {:[P(u","v","w)=P_(alpha beta gamma)omega^(alpha)(u)omega^(beta)(v)omega^(gamma)(w)],[(31.35)=P_(alpha beta gamma)u^(alpha)v^(beta)w^(gamma)]:}\begin{align*} \boldsymbol{P}(\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}) & =P_{\alpha \beta \gamma} \boldsymbol{\omega}^{\alpha}(\boldsymbol{u}) \boldsymbol{\omega}^{\beta}(\boldsymbol{v}) \boldsymbol{\omega}^{\gamma}(\boldsymbol{w}) \\ & =P_{\alpha \beta \gamma} u^{\alpha} v^{\beta} w^{\gamma} \tag{31.35} \end{align*}P(u,v,w)=Pαβγωα(u)ωβ(v)ωγ(w)(31.35)=Pαβγuαvβwγ
Here is a list of things one can do with a tensor.
  • Take the gradient. The gradient operation grad\boldsymbol{\nabla} adds an extra ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1) slot to the tensor. Therefore, P P grad P\boldsymbol{\nabla} \boldsymbol{P}P has four slots that accept vectors. We interpret this as meaning
(31.36) P ( u , v , w , z ) ( P ( u , v , w ) at tip of z P ( u , v , w ) at base of z ) . (31.36) P ( u , v , w , z ) ( P ( u , v , w )  at tip of  z P ( u , v , w )  at base of  z ) . {:(31.36)grad P(u","v","w","z)~~((P(u,v,w)" at tip of "z)/(-P(u,v,w)" at base of "z)).:}\begin{equation*} \boldsymbol{\nabla} P(\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}, \boldsymbol{z}) \approx\binom{\boldsymbol{P}(\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}) \text { at tip of } \boldsymbol{z}}{-\boldsymbol{P}(\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}) \text { at base of } \boldsymbol{z}} . \tag{31.36} \end{equation*}(31.36)P(u,v,w,z)(P(u,v,w) at tip of zP(u,v,w) at base of z).
In flat space, we write P ( u , v , w , z ) P ( u , v , w , z ) grad P(u,v,w,z)\boldsymbol{\nabla} \boldsymbol{P}(\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}, \boldsymbol{z})P(u,v,w,z) in component form as
(31.37) P α β γ x μ u α v β w γ z μ = P α β γ , μ u α v β w γ z μ (31.37) P α β γ x μ u α v β w γ z μ = P α β γ , μ u α v β w γ z μ {:(31.37)(delP_(alpha beta gamma))/(delx^(mu))u^(alpha)v^(beta)w^(gamma)z^(mu)=P_(alpha beta gamma,mu)u^(alpha)v^(beta)w^(gamma)z^(mu):}\begin{equation*} \frac{\partial P_{\alpha \beta \gamma}}{\partial x^{\mu}} u^{\alpha} v^{\beta} w^{\gamma} z^{\mu}=P_{\alpha \beta \gamma, \mu} u^{\alpha} v^{\beta} w^{\gamma} z^{\mu} \tag{31.37} \end{equation*}(31.37)Pαβγxμuαvβwγzμ=Pαβγ,μuαvβwγzμ
  • Contraction reduces the rank of a tensor by two. We do this by inserting a basis vector and basis 1 -form and sum over corresponding components. Let's do this to the ( 1 , 3 ) ( 1 , 3 ) (1,3)(1,3)(1,3) tensor S S S\boldsymbol{S}S, defined as
(31.38) S = S α β γ μ e μ ω α ω β ω γ (31.38) S = S α β γ μ e μ ω α ω β ω γ {:(31.38)S=S_(alpha beta gamma)^(mu)e_(mu)oxomega^(alpha)oxomega^(beta)oxomega^(gamma):}\begin{equation*} \boldsymbol{S}=S_{\alpha \beta \gamma}^{\mu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{\omega}^{\alpha} \otimes \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{\omega}^{\gamma} \tag{31.38} \end{equation*}(31.38)S=Sαβγμeμωαωβωγ
to make a ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor Q Q QQQ with components Q α γ Q α γ Q_(alpha gamma)Q_{\alpha \gamma}Qαγ. We can write an expression for a scalar given by
(31.39) Q ( u , v ) = S ( ω μ , u , e μ , v ) (31.39) Q ( u , v ) = S ω μ , u , e μ , v {:(31.39)Q(u","v)=S(omega^(mu),u,e_(mu),v):}\begin{equation*} \boldsymbol{Q}(\boldsymbol{u}, \boldsymbol{v})=\boldsymbol{S}\left(\boldsymbol{\omega}^{\mu}, \boldsymbol{u}, \boldsymbol{e}_{\mu}, \boldsymbol{v}\right) \tag{31.39} \end{equation*}(31.39)Q(u,v)=S(ωμ,u,eμ,v)
where a sum over the repeated index is assumed. In components, this is written as
(31.40) Q ( u , v ) = Q α β u α v γ = S μ α μ γ u α v γ (31.40) Q ( u , v ) = Q α β u α v γ = S μ α μ γ u α v γ {:(31.40)Q(u","v)=Q_(alpha beta)u^(alpha)v^(gamma)=S^(mu)_(alpha mu gamma)u^(alpha)v^(gamma):}\begin{equation*} \boldsymbol{Q}(\boldsymbol{u}, \boldsymbol{v})=Q_{\alpha \beta} u^{\alpha} v^{\gamma}=S^{\mu}{ }_{\alpha \mu \gamma} u^{\alpha} v^{\gamma} \tag{31.40} \end{equation*}(31.40)Q(u,v)=Qαβuαvγ=Sμαμγuαvγ
The summing over a common up and down index is often called 'contracting an index'.
  • Divergence Contract the gradient with a slot on the tensor. Let's consider our ( 1 , 3 ) ( 1 , 3 ) (1,3)(1,3)(1,3) tensor S S S\boldsymbol{S}S again. We'll assume we're taking a divergence with respect to the first slot
S ( u , v , w ) = S ( ω μ , u , v , w , e μ ) (31.41) = S μ α β γ x μ u α v β w γ = S α β γ , μ μ u α v β w γ , S ( u , v , w ) = S ω μ , u , v , w , e μ (31.41) = S μ α β γ x μ u α v β w γ = S α β γ , μ μ u α v β w γ , {:[grad*S(u","v","w)=grad S(omega^(mu),u,v,w,e_(mu))],[(31.41)=(delS^(mu)_(alpha beta gamma))/(delx^(mu))*u^(alpha)v^(beta)w^(gamma)=S_(alpha beta gamma,mu)^(mu)u^(alpha)v^(beta)w^(gamma)","]:}\begin{align*} \boldsymbol{\nabla} \cdot \boldsymbol{S}(\boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}) & =\boldsymbol{\nabla} \boldsymbol{S}\left(\boldsymbol{\omega}^{\mu}, \boldsymbol{u}, \boldsymbol{v}, \boldsymbol{w}, \boldsymbol{e}_{\mu}\right) \\ & =\frac{\partial S^{\mu}{ }_{\alpha \beta \gamma}}{\partial x^{\mu}} \cdot u^{\alpha} v^{\beta} w^{\gamma}=S_{\alpha \beta \gamma, \mu}^{\mu} u^{\alpha} v^{\beta} w^{\gamma}, \tag{31.41} \end{align*}S(u,v,w)=S(ωμ,u,v,w,eμ)(31.41)=Sμαβγxμuαvβwγ=Sαβγ,μμuαvβwγ,
where the component version applies in flat space.
  • Transpose This swaps the contents of two slots
(31.42) P ( u , v , w ) P ( v , u , w ) (31.42) P ( u , v , w ) P ( v , u , w ) {:(31.42)P(u","v","w)rarr P(v","u","w):}\begin{equation*} P(u, v, w) \rightarrow P(v, u, w) \tag{31.42} \end{equation*}(31.42)P(u,v,w)P(v,u,w)
What this does to a tensor depends on the properties of that tensor. A tensor is symmetric if it is unaffected by a the transpose of two slots. 19 19 ^(19){ }^{19}19 If it is unaffected by the swap of any two slots it is totally symmetric
(31.43) P ( u , v , w ) = P ( v , u , w ) = P ( w , u , v ) = (31.43) P ( u , v , w ) = P ( v , u , w ) = P ( w , u , v ) = {:(31.43)P(u","v","w)=P(v","u","w)=P(w","u","v)=dots:}\begin{equation*} P(u, v, w)=P(v, u, w)=P(w, u, v)=\ldots \tag{31.43} \end{equation*}(31.43)P(u,v,w)=P(v,u,w)=P(w,u,v)=
A tensor is antisymmetric if the sign is reversed when two slots are swapped. 20 20 ^(20){ }^{20}20 It is totally antisymmetric if the sign is reversed if any two slots are swapped.
(31.44) P ( u , v , w ) = P ( v , u , w ) = + P ( w , u , v ) = (31.44) P ( u , v , w ) = P ( v , u , w ) = + P ( w , u , v ) = {:(31.44)P(u","v","w)=-P(v","u","w)=+P(w","u","v)=dots:}\begin{equation*} P(u, v, w)=-P(v, u, w)=+P(w, u, v)=\ldots \tag{31.44} \end{equation*}(31.44)P(u,v,w)=P(v,u,w)=+P(w,u,v)=
  • Symmetrization In components, the symmetrization of a ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor T T T\boldsymbol{T}T involves the following action on the components:
(31.45) T ( μ ν ) = 1 2 ! ( T μ ν + T ν μ ) (31.45) T ( μ ν ) = 1 2 ! T μ ν + T ν μ {:(31.45)T^((mu nu))=(1)/(2!)(T^(mu nu)+T^(nu mu)):}\begin{equation*} T^{(\mu \nu)}=\frac{1}{2!}\left(T^{\mu \nu}+T^{\nu \mu}\right) \tag{31.45} \end{equation*}(31.45)T(μν)=12!(Tμν+Tνμ)
A symmetric tensor then has the property T ( μ ν ) = T μ ν T ( μ ν ) = T μ ν T^((mu nu))=T^(mu nu)T^{(\mu \nu)}=T^{\mu \nu}T(μν)=Tμν. An antisymmetric tensor has the property A ( μ ν ) = 0 A ( μ ν ) = 0 A^((mu nu))=0A^{(\mu \nu)}=0A(μν)=0. Symmetrization therefore extracts the symmetric part of a tensor. For a ( n , m n , m n,mn, mn,m ) tensor we have the rule
(31.46) T μ σ ( α κ ) = 1 n ! ( Sum over all permutations of the n indices α κ ) . (31.46) T μ σ ( α κ ) = 1 n ! (  Sum over all permutations   of the  n  indices  α κ ) . {:(31.46)T_(mu dots sigma)^((alpha dots kappa))=(1)/(n!)((" Sum over all permutations ")/(" of the "n" indices "alpha dots kappa)).:}\begin{equation*} T_{\mu \ldots \sigma}^{(\alpha \ldots \kappa)}=\frac{1}{n!}\binom{\text { Sum over all permutations }}{\text { of the } n \text { indices } \alpha \ldots \kappa} . \tag{31.46} \end{equation*}(31.46)Tμσ(ακ)=1n!( Sum over all permutations  of the n indices ακ).
  • Antisymmetrization In components the antisymmetrization of a tensor involved the following action on the components:
(31.47) T [ μ ν ] = 1 2 ! ( T μ ν T ν μ ) (31.47) T [ μ ν ] = 1 2 ! T μ ν T ν μ {:(31.47)T^([mu nu])=(1)/(2!)(T^(mu nu)-T^(nu mu)):}\begin{equation*} T^{[\mu \nu]}=\frac{1}{2!}\left(T^{\mu \nu}-T^{\nu \mu}\right) \tag{31.47} \end{equation*}(31.47)T[μν]=12!(TμνTνμ)
A symmetric tensor has the property T [ μ ν ] = 0 T [ μ ν ] = 0 T^([mu nu])=0T^{[\mu \nu]}=0T[μν]=0. An antisymmetric tensor has the property A [ μ ν ] = A μ ν A [ μ ν ] = A μ ν A^([mu nu])=A^(mu nu)A^{[\mu \nu]}=A^{\mu \nu}A[μν]=Aμν. Antisymmetrization therefore extracts the antisymmetric part of a tensor. For a ( n , m ) ( n , m ) (n,m)(n, m)(n,m) tensor we have the rule for antisymmetrization that
(31.48) T μ σ [ α κ ] = 1 n ! ( Alternating sum over all permutations of the n indices α κ ) . (31.48) T μ σ [ α κ ] = 1 n ! (  Alternating sum over all   permutations of the  n  indices  α κ ) . {:(31.48)T_(mu dots sigma)^([alpha dots kappa])=(1)/(n!)((" Alternating sum over all ")/(" permutations of the "n" indices "alpha dots kappa)).:}\begin{equation*} T_{\mu \ldots \sigma}^{[\alpha \ldots \kappa]}=\frac{1}{n!}\binom{\text { Alternating sum over all }}{\text { permutations of the } n \text { indices } \alpha \ldots \kappa} . \tag{31.48} \end{equation*}(31.48)Tμσ[ακ]=1n!( Alternating sum over all  permutations of the n indices ακ).
  • The wedge product is an antisymmetrized tensor product. Take vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v and build the bivector b b b\boldsymbol{b}b using the wedge product
(31.49) b = u v = u v v u (31.49) b = u v = u v v u {:(31.49)b=u^^v=u ox v-v ox u:}\begin{equation*} \boldsymbol{b}=\boldsymbol{u} \wedge \boldsymbol{v}=\boldsymbol{u} \otimes \boldsymbol{v}-\boldsymbol{v} \otimes \boldsymbol{u} \tag{31.49} \end{equation*}(31.49)b=uv=uvvu
The components of the bivector are
b μ ν = b ( ω μ , ω ν ) = u [ ω μ ] v [ ω ν ] v [ ω μ ] u [ ω ν ] (31.50) = u μ v ν v μ u ν b μ ν = b ω μ , ω ν = u ω μ v ω ν v ω μ u ω ν (31.50) = u μ v ν v μ u ν {:[b^(mu nu)=b(omega^(mu),omega^(nu))=u[omega^(mu)]ox v[omega^(nu)]-v[omega^(mu)]ox u[omega^(nu)]],[(31.50)=u^(mu)v^(nu)-v^(mu)u^(nu)]:}\begin{align*} b^{\mu \nu}=\boldsymbol{b}\left(\boldsymbol{\omega}^{\mu}, \boldsymbol{\omega}^{\nu}\right) & =\boldsymbol{u}\left[\boldsymbol{\omega}^{\mu}\right] \otimes \boldsymbol{v}\left[\boldsymbol{\omega}^{\nu}\right]-\boldsymbol{v}\left[\boldsymbol{\omega}^{\mu}\right] \otimes \boldsymbol{u}\left[\boldsymbol{\omega}^{\nu}\right] \\ & =u^{\mu} v^{\nu}-v^{\mu} u^{\nu} \tag{31.50} \end{align*}bμν=b(ωμ,ων)=u[ωμ]v[ων]v[ωμ]u[ων](31.50)=uμvνvμuν
19 19 ^(19){ }^{19}19 If a ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor is symmetric then the components obey T μ ν = T ν μ T μ ν = T ν μ T^(mu nu)=T^(nu mu)T^{\mu \nu}=T^{\nu \mu}Tμν=Tνμ.
20 20 ^(20){ }^{20}20 If a ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor is antisymmetric then the components obey A μ ν = A μ ν = A^(mu nu)=A^{\mu \nu}=Aμν= A ν μ A ν μ -A^(nu mu)-A^{\nu \mu}Aνμ.
We shall use these tricks in the chapters that follow.
We now have a set of vectors, 1 -forms and tensors with which to describe geometry. They can be written in a coordinate-independent form (i.e. v , σ ~ , T v , σ ~ , T v, tilde(sigma),T\boldsymbol{v}, \tilde{\boldsymbol{\sigma}}, \boldsymbol{T}v,σ~,T ) or we can write them in terms of a set of coordinates by invoking basis vectors and components. By combining the objects using the relevant slots we can map between them and numbers. Each of these objects is useful in describing physical quantities commonly found in Nature.
In the next chapter, we will treat a special class of tensor that are created from a special combination of 1-forms. These are called differential forms and are of particular importance owing to their use in describing curvature.

Chapter summary

  • Vectors can be identified with derivatives. The tangent vector at a point parametrized by λ λ lambda\lambdaλ on a curve P ( λ ) P ( λ ) P(lambda)\mathcal{P}(\lambda)P(λ) is v = d / d λ v = d / d λ v=d//dlambda\boldsymbol{v}=\mathrm{d} / \mathrm{d} \lambdav=d/dλ. A vector field allows us to find a vector at each point.
  • A vector can be combined with a 1 -form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ to make a number. A 1 -form field gives a 1 -form at every point in space. It only makes sense to combine vectors and 1 -forms defined at the same point.
  • Tensors generalize the notion of vectors and 1 -forms. Tensors can be manipulated in a large number of ways to produce tensors of different valences.

Exercises

(31.1) The basis vectors and basis 1-forms for a coordinate frame are given by / x μ / x μ del//delx^(mu)\partial / \partial x^{\mu}/xμ and d x α d x α dx^(alpha)\boldsymbol{d} x^{\alpha}dxα, respectively. Evaluate:
(a) d x 1 , / x 1 d x 1 , / x 1 (:dx^(1),del//delx^(1):)\left\langle\boldsymbol{d} x^{1}, \partial / \partial x^{1}\right\rangledx1,/x1,
(b) d x 1 , / x 0 d x 1 , / x 0 (:dx^(1),del//delx^(0):)\left\langle\boldsymbol{d} x^{1}, \partial / \partial x^{0}\right\rangledx1,/x0,
(c) / x α / x β / x α / x β del//delx^(alpha)*del//delx^(beta)\partial / \partial x^{\alpha} \cdot \partial / \partial x^{\beta}/xα/xβ,
(d) d x α d x β d x α d x β dx^(alpha)*dx^(beta)\boldsymbol{d} x^{\alpha} \cdot \boldsymbol{d} x^{\beta}dxαdxβ.
(31.2) Find a 1 -form W ~ W ~ tilde(W)\tilde{\boldsymbol{W}}W~ that acts on a displacement X X vec(X)\vec{X}X to give the work done by a force f f vec(f)\vec{f}f in Euclidean space.
(31.3) Expand (a) T ( α β γ ) δ ϵ ς T ( α β γ ) δ ϵ ς T^((alpha beta gamma))_(delta epsilonς)T^{(\alpha \beta \gamma)}{ }_{\delta \epsilon \varsigma}T(αβγ)δϵς; (b) T α β γ [ δ ϵ ζ ] T α β γ [ δ ϵ ζ ] T^(alpha beta gamma)_([delta epsilon zeta])T^{\alpha \beta \gamma}{ }_{[\delta \epsilon \zeta]}Tαβγ[δϵζ].
(31.4) Consider the contraction A μ ν T μ ν A μ ν T μ ν A^(mu nu)T_(mu nu)A^{\mu \nu} T_{\mu \nu}AμνTμν, where T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν are the components of some general tensor.
(a) Show that if A μ ν A μ ν A^(mu nu)A^{\mu \nu}Aμν is symmetric, only the symmetric part of T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν contributes to the contraction.
(b) Show that if A μ ν A μ ν A^(mu nu)A^{\mu \nu}Aμν is antisymmetric, only the asymmetric part of T μ ν T μ ν T_(mu nu)T_{\mu \nu}Tμν contributes to the contraction.
(31.5) (a) Expand F [ μ ν ; λ ] F [ μ ν ; λ ] F_([mu nu;lambda])F_{[\mu \nu ; \lambda]}F[μν;λ] if F μ ν F μ ν F_(mu nu)F_{\mu \nu}Fμν is an antisymmetric tensor
(b) Show that
(31.51) F μ ν η μ α η ν β u α u β = 0 (31.51) F μ ν η μ α η ν β u α u β = 0 {:(31.51)F^(mu nu)eta_(mu alpha)eta_(nu beta)u^(alpha)u^(beta)=0:}\begin{equation*} F^{\mu \nu} \eta_{\mu \alpha} \eta_{\nu \beta} u^{\alpha} u^{\beta}=0 \tag{31.51} \end{equation*}(31.51)Fμνημαηνβuαuβ=0
where F μ ν F μ ν F^(mu nu)F^{\mu \nu}Fμν are the components of an antisymmetric tensor.
(31.6) By removing the point ξ μ ξ μ xi^(mu)\xi^{\mu}ξμ, show that the expression
(31.52) 2 ξ μ x λ x σ = Γ λ σ ν ξ μ x ν (31.52) 2 ξ μ x λ x σ = Γ λ σ ν ξ μ x ν {:(31.52)(del^(2)xi^(mu))/(delx^(lambda)delx^(sigma))=Gamma_(lambda sigma)^(nu)(delxi^(mu))/(delx^(nu)):}\begin{equation*} \frac{\partial^{2} \xi^{\mu}}{\partial x^{\lambda} \partial x^{\sigma}}=\Gamma_{\lambda \sigma}^{\nu} \frac{\partial \xi^{\mu}}{\partial x^{\nu}} \tag{31.52} \end{equation*}(31.52)2ξμxλxσ=Γλσνξμxν
is equivalent to
(31.53) e σ x λ = Γ λ σ ν e ν (31.53) e σ x λ = Γ λ σ ν e ν {:(31.53)(dele_(sigma))/(delx^(lambda))=Gamma_(lambda sigma)^(nu)e_(nu):}\begin{equation*} \frac{\partial \boldsymbol{e}_{\sigma}}{\partial x^{\lambda}}=\Gamma_{\lambda \sigma}^{\nu} \boldsymbol{e}_{\nu} \tag{31.53} \end{equation*}(31.53)eσxλ=Γλσνeν
(31.7) Roger Penrose has invented a diagrammatic notation for tensor equations that provides an entertaining way of visualizing complicated expressions. Tensors are represented by shapes. The rules are that a valence ( n , m ) ( n , m ) (n,m)(n, m)(n,m) tensor has n n nnn lines emerging from its top and m m mmm lines from the bottom. The position of the page allows one to keep track of indices. A thick bar denotes antisymmetrization and a wiggly bar denotes symmetrization (although the numerical factors associated with these
operations are not conventionally included, so can be drawn on the diagrams).
(a) Using these rules and referring to Fig. 31.10, explain why diagram (i) represents the component expression A μ ν α β γ + 2 A μ ν β γ α A μ ν α β γ + 2 A μ ν β γ α A^(mu nu)_(alpha beta gamma)+2A^(mu nu)_(beta gamma alpha)A^{\mu \nu}{ }_{\alpha \beta \gamma}+2 A^{\mu \nu}{ }_{\beta \gamma \alpha}Aμναβγ+2Aμνβγα.
(b) What expression does diagram (ii) represent? What about (iii)?
(c) If covariant derivatives are represented by a large circle around the tensor, suggest which equations of electromagnetism are represented by diagrams (iv) and (v).
(d) Which expression involving the Riemann tensor is represented by diagram (vi)?
See Penrose (2004) for more details.
(i)

(ii)

(iii)

(iv) = = =◻=\square=
(v)

(vi)
Fig. 31.10 Tensor diagrams from Exercise 31.7.
32
32.1 2-forms
32.2 -forms 334
32.3 -vectors 336
Chapter summary 337
Exercises 339
32.1 2-forms 32.2 -forms 334 32.3 -vectors 336 Chapter summary 337 Exercises 339| 32.1 | 2-forms | | :--- | :--- | | 32.2 -forms | 334 | | 32.3 -vectors | 336 | | Chapter summary | 337 | | Exercises | 339 |
Fig. 32.1 An example 2-form β ~ ( β ~ ( tilde(beta)(\tilde{\boldsymbol{\beta}}(β~(, ) made from ω x ω y ω x ω y omega^(x)^^omega^(y)\boldsymbol{\omega}^{x} \wedge \boldsymbol{\omega}^{y}ωxωy.
1 1 ^(1){ }^{1}1 There's an analogy here between the vectors and 1 -forms that we have described, with their characteristic relascribed, with their characteristic rela-
tionship ω μ , e μ = δ μ ν ω μ , e μ = δ μ ν (:omega^(mu),e_(mu):)=delta^(mu)_(nu)\left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{e}_{\mu}\right\rangle=\delta^{\mu}{ }_{\nu}ωμ,eμ=δμν, and the idea tionship ω μ , e μ = δ ν ω μ , e μ = δ ν (:omega^(mu),e_(mu):)=delta_(nu)\left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{e}_{\mu}\right\rangle=\delta_{\nu}ωμ,eμ=δν, , and the idea
of lattice vectors a i a i a_(i)\boldsymbol{a}_{i}ai and reciprocal latof lattice vectors a i a i a_(i)\boldsymbol{a}_{i}ai and reciprocal lat-
tice vectors A j A j A_(j)\boldsymbol{A}_{j}Aj in solids, which have an tice vectors A j A j A_(j)\boldsymbol{A}_{j}Aj in solids, which have an
analogous flat-space relationship A i A i A_(i)\boldsymbol{A}_{i}Ai. analogous flat-space relationship A i A i A_(i)\boldsymbol{A}_{i}Ai
a j = 2 π δ i j a j = 2 π δ i j a_(j)=2pidelta_(ij)\boldsymbol{a}_{j}=2 \pi \delta_{i j}aj=2πδij. We can use reciprocal lattice vectors to label sets of lattice planes, just as we label repeating planes using 1 -forms. See the book by Pauli, Section 10.
2 2 ^(2){ }^{2}2 As we saw in the last chapter, we can also use the wedge product on vectors like u u u\boldsymbol{u}u and v v v\boldsymbol{v}v to make bivectors
u v = u v v u u v = u v v u u^^v=u ox v-v ox u\boldsymbol{u} \wedge \boldsymbol{v}=\boldsymbol{u} \otimes \boldsymbol{v}-\boldsymbol{v} \otimes \boldsymbol{u}uv=uvvu
Bivectors are antisymmetric ( 2,0 ) objects with two slots, each of which takes a 1 -form.

Differential forms

Good God, there's two of them!

Audience heckle (on seeing the comic Bernie Winters take the stage at the Glasgow Empire, after his brother Mike had been poorly received)
In the last chapter, we reacquainted ourselves with vectors and 1forms. With certain qualifications, the former can be thought of as arrows; the latter as a set of regularly repeating surfaces. In this chapter, we start with the 1 -form and use it to generate a family of objects known as differential forms or p p ppp-forms. These have several applications in general relativity: a particularly useful example is that curvature can be described very efficiently as an object known as a 2 2 2\mathbf{2}2-form. We will see how to encode information about areas and volumes in forms, and this will lead us, in Chapter 38, to the profound idea that every integral can be reinterpreted as an integral over a form.

32.1 2-forms

1-forms are the simplest of a family of tensor objects called differential forms. This family of forms can be built from 1-forms. Recall the geometric interpretation of a 1 -form as an infinite number of equally spaced surfaces. 1 1 ^(1){ }^{1}1 By combining the surfaces of 1 -forms to build more complicated structures, we can generate all of the other differential forms.
Let's look at an example. Working in Cartesian coordinates, the surfaces of the basis 1-form ω μ ω μ omega^(mu)\boldsymbol{\omega}^{\mu}ωμ are perpendicular to the direction of the basis vector e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ. By combining the surfaces ω x ω x omega^(x)\boldsymbol{\omega}^{x}ωx and ω y ω y omega^(y)\boldsymbol{\omega}^{y}ωy as shown in Fig. 32.1 we can create the structure of tubes shown in the figure. This structure corresponds to an antisymmetric ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor β ~ ( β ~ ( tilde(beta)(\tilde{\boldsymbol{\beta}}(β~(, ) , w h i c h i s ) , w h i c h i s ),whichis) , which is),whichis our first example of a 2 2 2\mathbf{2}2-form. To achieve the combination of the surfaces symbolically, we use the wedge product to combine the 1 -forms. 2 2 ^(2){ }^{2}2 The wedge product of two 1 -forms, σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ and τ ~ τ ~ tilde(tau)\tilde{\boldsymbol{\tau}}τ~, is the antisymmetrized tensor product
(32.2) σ ~ τ ~ = σ ~ τ ~ τ ~ σ ~ (32.2) σ ~ τ ~ = σ ~ τ ~ τ ~ σ ~ {:(32.2) tilde(sigma)^^ tilde(tau)= tilde(sigma)ox tilde(tau)- tilde(tau)ox tilde(sigma):}\begin{equation*} \tilde{\boldsymbol{\sigma}} \wedge \tilde{\boldsymbol{\tau}}=\tilde{\boldsymbol{\sigma}} \otimes \tilde{\boldsymbol{\tau}}-\tilde{\boldsymbol{\tau}} \otimes \tilde{\boldsymbol{\sigma}} \tag{32.2} \end{equation*}(32.2)σ~τ~=σ~τ~τ~σ~
The wedge product to two 1 -forms is a 2 -form.
We notice immediately that with this definition
(32.3) σ ~ σ ~ = 0 , (32.3) σ ~ σ ~ = 0 , {:(32.3) tilde(sigma)^^ tilde(sigma)=0",":}\begin{equation*} \tilde{\boldsymbol{\sigma}} \wedge \tilde{\boldsymbol{\sigma}}=0, \tag{32.3} \end{equation*}(32.3)σ~σ~=0,
where σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ is any 1 -form.
With this notation, our example 2-form β ~ ( β ~ ( tilde(beta)(\tilde{\boldsymbol{\beta}}(β~(, ) c a n b e g e n e r a t e d b y ) c a n b e g e n e r a t e d b y )canbegeneratedby) can be generated by)canbegeneratedby saying
(32.4) β ~ ( , ) = ω x ω y . (32.4) β ~ ( , ) = ω x ω y . {:(32.4) tilde(beta)(",")=omega^(x)^^omega^(y).:}\begin{equation*} \tilde{\boldsymbol{\beta}}(,)=\boldsymbol{\omega}^{x} \wedge \boldsymbol{\omega}^{y} . \tag{32.4} \end{equation*}(32.4)β~(,)=ωxωy.
The wedge-product operation does not generate an arbitrary ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor; rather, it is antisymmetric. 3 3 ^(3){ }^{3}3 The antisymmetry is interpreted geometrically in terms of the tubes in Fig. 32.1 having a handedness. To visualize this, we could picture the 2 -form circulating in a particular direction in each tube as shown in Fig. 32.2.
Let's now consider a more general 2-form B ~ ( ) = , σ ~ τ ~ B ~ ( ) = , σ ~ τ ~ tilde(B)()=, tilde(sigma)^^ tilde(tau)\tilde{\boldsymbol{B}}()=,\tilde{\boldsymbol{\sigma}} \wedge \tilde{\boldsymbol{\tau}}B~()=,σ~τ~, built from two arbitrary 1-forms σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ and τ ~ τ ~ tilde(tau)\tilde{\boldsymbol{\tau}}τ~. Any 2 -form can be written in terms of its components B μ ν B μ ν B_(mu nu)B_{\mu \nu}Bμν as
(32.5) B ~ ( , ) = 1 2 B μ ν ω μ ω ν , (32.5) B ~ ( , ) = 1 2 B μ ν ω μ ω ν , {:(32.5) tilde(B)(",")=(1)/(2)B_(mu nu)omega^(mu)^^omega^(nu)",":}\begin{equation*} \tilde{\boldsymbol{B}}(,)=\frac{1}{2} B_{\mu \nu} \boldsymbol{\omega}^{\mu} \wedge \boldsymbol{\omega}^{\nu}, \tag{32.5} \end{equation*}(32.5)B~(,)=12Bμνωμων,
where ω μ ω μ omega^(mu)\boldsymbol{\omega}^{\mu}ωμ are basis 1 -forms. The asymmetry of the 2 -form means that B μ ν = B ν μ B μ ν = B ν μ B_(mu nu)=-B_(nu mu)B_{\mu \nu}=-B_{\nu \mu}Bμν=Bνμ. The factor 1 / 2 1 / 2 1//21 / 21/2 is a convention, that we justify in the next example. 4 4 ^(4){ }^{4}4

Example 32.1

The motivation for the 1 / 2 1 / 2 1//21 / 21/2 factor is seen by expanding the 2 -form
(32.8) B ~ = 1 2 B μ ν ω μ ω ν = 1 2 B μ ν ( ω μ ω ν ω ν ω μ ) (32.8) B ~ = 1 2 B μ ν ω μ ω ν = 1 2 B μ ν ω μ ω ν ω ν ω μ {:[(32.8) tilde(B)=(1)/(2)B_(mu nu)omega^(mu)^^omega^(nu)],[=(1)/(2)B_(mu nu)(omega^(mu)oxomega^(nu)-omega^(nu)oxomega^(mu))]:}\begin{align*} \tilde{\boldsymbol{B}} & =\frac{1}{2} B_{\mu \nu} \boldsymbol{\omega}^{\mu} \wedge \boldsymbol{\omega}^{\nu} \tag{32.8}\\ & =\frac{1}{2} B_{\mu \nu}\left(\boldsymbol{\omega}^{\mu} \otimes \boldsymbol{\omega}^{\nu}-\boldsymbol{\omega}^{\nu} \otimes \boldsymbol{\omega}^{\mu}\right) \end{align*}(32.8)B~=12Bμνωμων=12Bμν(ωμωνωνωμ)
Now insert basis vectors e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα and e β e β e_(beta)\boldsymbol{e}_{\beta}eβ to extract the components of the 2-form
B ~ ( e α , e β ) = B μ ν 2 ( δ μ α δ ν β δ ν α δ μ β ) (32.9) = 1 2 ( B α β B β α ) = B α β , B ~ e α , e β = B μ ν 2 δ μ α δ ν β δ ν α δ μ β (32.9) = 1 2 B α β B β α = B α β , {:[ tilde(B)(e_(alpha),e_(beta))=(B_(mu nu))/(2)*(delta^(mu)_(alpha)delta^(nu)_(beta)-delta^(nu)_(alpha)delta^(mu)_(beta))],[(32.9)=(1)/(2)(B_(alpha beta)-B_(beta alpha))=B_(alpha beta)","]:}\begin{align*} \tilde{\boldsymbol{B}}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right) & =\frac{B_{\mu \nu}}{2} \cdot\left(\delta^{\mu}{ }_{\alpha} \delta^{\nu}{ }_{\beta}-\delta^{\nu}{ }_{\alpha}{\delta^{\mu}}_{\beta}\right) \\ & =\frac{1}{2}\left(B_{\alpha \beta}-B_{\beta \alpha}\right)=B_{\alpha \beta}, \tag{32.9} \end{align*}B~(eα,eβ)=Bμν2(δμαδνβδναδμβ)(32.9)=12(BαβBβα)=Bαβ,
where, in the final step we use asymmetry to say B α β = B β α B α β = B β α B_(alpha beta)=-B_(beta alpha)B_{\alpha \beta}=-B_{\beta \alpha}Bαβ=Bβα.
Just as with the tensors from the last chapter, it is possible to use the slot-machine structure of forms to extract their components.

Example 32.2

We can extract the components of the 2 -form B ~ = σ ~ τ ~ B ~ = σ ~ τ ~ tilde(B)= tilde(sigma)^^ tilde(tau)\tilde{\boldsymbol{B}}=\tilde{\boldsymbol{\sigma}} \wedge \tilde{\boldsymbol{\tau}}B~=σ~τ~ in terms of the components of σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ and τ ~ τ ~ tilde(tau)\tilde{\boldsymbol{\tau}}τ~ by inserting basis vectors e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ into the slots in the expression B ~ ( ) = σ ~ ( ) τ ~ ( ) B ~ ( ) = σ ~ ( ) τ ~ ( ) tilde(B)()= tilde(sigma)()^^ tilde(tau)()\tilde{\boldsymbol{B}}()=\tilde{\boldsymbol{\sigma}}() \wedge \tilde{\boldsymbol{\tau}}()B~()=σ~()τ~() to find
B μ ν = B ~ ( e μ , e ν ) = σ ~ ( e μ ) τ ~ ( e ν ) τ ~ ( e μ ) σ ~ ( e ν ) = σ μ τ ν τ μ σ ν B μ ν = B ~ e μ , e ν = σ ~ e μ τ ~ e ν τ ~ e μ σ ~ e ν = σ μ τ ν τ μ σ ν {:[B_(mu nu)= tilde(B)(e_(mu),e_(nu))],[= tilde(sigma)(e_(mu))ox tilde(tau)(e_(nu))- tilde(tau)(e_(mu))ox tilde(sigma)(e_(nu))],[=sigma_(mu)tau_(nu)-tau_(mu)sigma_(nu)]:}\begin{aligned} B_{\mu \nu} & =\tilde{\boldsymbol{B}}\left(\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right) \\ & =\tilde{\boldsymbol{\sigma}}\left(\boldsymbol{e}_{\mu}\right) \otimes \tilde{\boldsymbol{\tau}}\left(\boldsymbol{e}_{\nu}\right)-\tilde{\boldsymbol{\tau}}\left(\boldsymbol{e}_{\mu}\right) \otimes \tilde{\boldsymbol{\sigma}}\left(\boldsymbol{e}_{\nu}\right) \\ & =\sigma_{\mu} \tau_{\nu}-\tau_{\mu} \sigma_{\nu} \end{aligned}Bμν=B~(eμ,eν)=σ~(eμ)τ~(eν)τ~(eμ)σ~(eν)=σμτντμσν
3 3 ^(3){ }^{3}3 Recall from the previous chapter that a ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor is antisymmetric if B ~ ( x , y ) = B ( y , x ) B ~ ( x , y ) = B ( y , x ) tilde(B)(x,y)=-B(y,x)\tilde{B}(\boldsymbol{x}, \boldsymbol{y})=-\boldsymbol{B}(\boldsymbol{y}, \boldsymbol{x})B~(x,y)=B(y,x), where x x x\boldsymbol{x}x and y y y\boldsymbol{y}y are vectors.
Fig. 32.2 The 1 -forms ω x ω x omega^(x)\boldsymbol{\omega}^{x}ωx and ω y ω y omega^(y)\boldsymbol{\omega}^{y}ωy can be thought of as sets of parallel surfaces. Their combination gives the tubular structure described by ω x ω y ω x ω y omega^(x)^^omega^(y)\boldsymbol{\omega}^{x} \wedge \boldsymbol{\omega}^{y}ωxωy, with a particular handedness, as depicted by the arrows in the tubes.
4 4 ^(4){ }^{4}4 As shown in the example, the factor 1 / 2 1 / 2 1//21 / 21/2 only arises when we're summing over indices, as in eqn 32.5. If we choose to specify specific components, rather than sum over them, no factor explicit factor is needed. For example, if a 2 form in Euclidean 3-space has only one non-zero component, we write
(32.6) B ~ = γ ( ω 2 ω 3 ) (32.6) B ~ = γ ω 2 ω 3 {:(32.6) tilde(B)=gamma(omega^(2)^^omega^(3)):}\begin{equation*} \tilde{\boldsymbol{B}}=\gamma\left(\boldsymbol{\omega}^{2} \wedge \boldsymbol{\omega}^{3}\right) \tag{32.6} \end{equation*}(32.6)B~=γ(ω2ω3)
Computing components we find
B μ ν = γ [ ω 2 ( e μ ) ω 3 ( e ν ) B μ ν = γ ω 2 e μ ω 3 e ν B_(mu nu)=gamma[omega^(2)(e_(mu))oxomega^(3)(e_(nu)):}B_{\mu \nu}=\gamma\left[\boldsymbol{\omega}^{2}\left(\boldsymbol{e}_{\mu}\right) \otimes \boldsymbol{\omega}^{3}\left(\boldsymbol{e}_{\nu}\right)\right.Bμν=γ[ω2(eμ)ω3(eν)
(32.7) ω 3 ( e μ ) ω 2 ( e ν ) (32.7) ω 3 e μ ω 2 e ν {:(32.7)-omega^(3)(e_(mu))oxomega^(2)(e_(nu)):}\begin{equation*} -\omega^{3}\left(e_{\mu}\right) \otimes \omega^{2}\left(e_{\nu}\right) \tag{32.7} \end{equation*}(32.7)ω3(eμ)ω2(eν)
= γ [ δ μ 2 δ ν 3 δ μ 3 δ ν 2 ] = γ δ μ 2 δ ν 3 δ μ 3 δ ν 2 =gamma[delta_(mu)^(2)delta_(nu)^(3)-delta_(mu)^(3)delta_(nu)^(2)]=\gamma\left[\delta_{\mu}^{2} \delta_{\nu}^{3}-\delta_{\mu}^{3} \delta_{\nu}^{2}\right]=γ[δμ2δν3δμ3δν2],
so B 23 = γ B 23 = γ B_(23)=gammaB_{23}=\gammaB23=γ and B 32 = γ B 32 = γ B_(32)=-gammaB_{32}=-\gammaB32=γ.
5 5 ^(5){ }^{5}5 Recall from the last chapter that the spacing of the surfaces in the Cartesian μ μ mu\muμ-direction is 1 / σ μ 1 / σ μ 1//sigma_(mu)1 / \sigma_{\mu}1/σμ.
Fig. 32.3 An arbitrary 2-form.
6 6 ^(6){ }^{6}6 We define a 0 -form to be a function such as f ( x ) f ( x ) f(x)f(x)f(x).
7 7 ^(7){ }^{7}7 For example in three-dimensional space
B ~ = 1 2 B μ ν ( ω μ ω ν ) = 1 2 ( B 12 ω 1 ω 2 + B 21 ω 2 ω 1 + B 13 ω 1 ω 3 + B 31 ω 3 ω 1 + B 23 ω 2 ω 3 + B 32 ω 3 ω 2 ) . B ~ = 1 2 B μ ν ω μ ω ν = 1 2 B 12 ω 1 ω 2 + B 21 ω 2 ω 1 + B 13 ω 1 ω 3 + B 31 ω 3 ω 1 + B 23 ω 2 ω 3 + B 32 ω 3 ω 2 . {:[ tilde(B)=(1)/(2)B_(mu nu)(omega^(mu)^^omega^(nu))],[=(1)/(2)(B_(12)omega^(1)^^omega^(2)+B_(21)omega^(2)^^omega^(1):}],[+B_(13)omega^(1)^^omega^(3)+B_(31)omega^(3)^^omega^(1)],[{:+B_(23)omega^(2)^^omega^(3)+B_(32)omega^(3)^^omega^(2)).]:}\begin{aligned} \tilde{\boldsymbol{B}}= & \frac{1}{2} B_{\mu \nu}\left(\boldsymbol{\omega}^{\mu} \wedge \boldsymbol{\omega}^{\nu}\right) \\ = & \frac{1}{2}\left(B_{12} \boldsymbol{\omega}^{1} \wedge \boldsymbol{\omega}^{2}+B_{21} \boldsymbol{\omega}^{2} \wedge \boldsymbol{\omega}^{1}\right. \\ & +B_{13} \boldsymbol{\omega}^{1} \wedge \boldsymbol{\omega}^{3}+B_{31} \boldsymbol{\omega}^{3} \wedge \boldsymbol{\omega}^{1} \\ & \left.+B_{23} \boldsymbol{\omega}^{2} \wedge \boldsymbol{\omega}^{3}+B_{32} \boldsymbol{\omega}^{3} \wedge \boldsymbol{\omega}^{2}\right) . \end{aligned}B~=12Bμν(ωμων)=12(B12ω1ω2+B21ω2ω1+B13ω1ω3+B31ω3ω1+B23ω2ω3+B32ω3ω2).
Using (i) B μ ν = B ν μ B μ ν = B ν μ B_(mu nu)=-B_(nu mu)B_{\mu \nu}=-B_{\nu \mu}Bμν=Bνμ and (ii) ω μ ω μ omega^(mu)^^\boldsymbol{\omega}^{\mu} \wedgeωμ ω ν = ω ν ω μ ω ν = ω ν ω μ omega^(nu)=-omega^(nu)^^omega^(mu)\boldsymbol{\omega}^{\nu}=-\boldsymbol{\omega}^{\nu} \wedge \boldsymbol{\omega}^{\mu}ων=ωνωμ, we can simplify this to read B ~ = 1 2 B μ ν ( ω μ ω ν ) B ~ = 1 2 B μ ν ω μ ω ν tilde(B)=(1)/(2)B_(mu nu)(omega^(mu)^^omega^(nu))\tilde{\boldsymbol{B}}=\frac{1}{2} B_{\mu \nu}\left(\boldsymbol{\omega}^{\mu} \wedge \boldsymbol{\omega}^{\nu}\right)B~=12Bμν(ωμων) = ( B 12 ω 1 ω 2 + B 13 ω 1 ω 3 = B 12 ω 1 ω 2 + B 13 ω 1 ω 3 =(B_(12)omega^(1)^^omega^(2)+B_(13)omega^(1)^^omega^(3):}=\left(B_{12} \boldsymbol{\omega}^{1} \wedge \boldsymbol{\omega}^{2}+B_{13} \boldsymbol{\omega}^{1} \wedge \boldsymbol{\omega}^{3}\right.=(B12ω1ω2+B13ω1ω3 + B 23 ω 2 ω 3 ) + B 23 ω 2 ω 3 {:+B_(23)omega^(2)^^omega^(3))\left.+B_{23} \omega^{2} \wedge \omega^{3}\right)+B23ω2ω3) = B | μ ν | ω μ ω ν = B | μ ν | ω μ ω ν =B_(|mu nu|)omega^(mu)^^omega^(nu)=B_{|\mu \nu|} \boldsymbol{\omega}^{\mu} \wedge \boldsymbol{\omega}^{\nu}=B|μν|ωμων.

Example 32.3

Let's try inserting a single vector u u u\boldsymbol{u}u into the first slot of a 2 -form. (Remember that the tensor products ox\otimes are intended to maintain the order of the linear slots.) We find
(32.11) B ~ ( u , ) = σ ~ τ ~ ( u , ) = σ ~ ( u ) τ ~ ( ) τ ~ ( u ) σ ~ ( ) (32.11) B ~ ( u , ) = σ ~ τ ~ ( u , ) = σ ~ ( u ) τ ~ ( ) τ ~ ( u ) σ ~ ( ) {:(32.11) tilde(B)(u",")= tilde(sigma)^^ tilde(tau)(u",")= tilde(sigma)(u)ox tilde(tau)()- tilde(tau)(u)ox tilde(sigma)():}\begin{equation*} \tilde{B}(u,)=\tilde{\sigma} \wedge \tilde{\tau}(u,)=\tilde{\sigma}(u) \otimes \tilde{\tau}()-\tilde{\tau}(u) \otimes \tilde{\boldsymbol{\sigma}}() \tag{32.11} \end{equation*}(32.11)B~(u,)=σ~τ~(u,)=σ~(u)τ~()τ~(u)σ~()
The result is a 1 -form: a ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1) tensor object with a single available slot that accepts a vector.
An arbitrary 1 -form σ ~ = σ μ ω μ σ ~ = σ μ ω μ tilde(sigma)=sigma_(mu)omega^(mu)\tilde{\boldsymbol{\sigma}}=\sigma_{\mu} \boldsymbol{\omega}^{\mu}σ~=σμωμ is represented by a series of surfaces oriented according to, and with spacing determined by, the components 5 5 ^(5){ }^{5}5. The product of two such arbitrary 1-forms resembles the example shown in Fig. 32.3 made of intersecting surfaces giving a series of (parallelepiped) tubes. As in our simple example earlier, the tubes come with a sense of direction since there is a difference between σ ~ τ ~ σ ~ τ ~ tilde(sigma)^^ tilde(tau)\tilde{\boldsymbol{\sigma}} \wedge \tilde{\boldsymbol{\tau}}σ~τ~ and τ ~ σ ~ τ ~ σ ~ tilde(tau)^^ tilde(sigma)\tilde{\boldsymbol{\tau}} \wedge \tilde{\boldsymbol{\sigma}}τ~σ~. We think of the 2 -form field as circulating in each of the cells.

32.2 -forms

We've seen that the wedge product of two 1 -forms is a 2 -form. In fact, we can generalize this notion and say that, if μ ~ μ ~ tilde(mu)\tilde{\boldsymbol{\mu}}μ~ is a p p ppp-form and ν ~ ν ~ tilde(nu)\tilde{\boldsymbol{\nu}}ν~ is a q q qqq-form, then we can use the wedge product to make a ( p + q ) ( p + q ) (p+q)(p+q)(p+q)-form μ ~ ν ~ μ ~ ν ~ tilde(mu)^^ tilde(nu)\tilde{\boldsymbol{\mu}} \wedge \tilde{\boldsymbol{\nu}}μ~ν~, with the property that
(32.12) μ ~ ν ~ = ( 1 ) p q ν ~ μ ~ (32.12) μ ~ ν ~ = ( 1 ) p q ν ~ μ ~ {:(32.12) tilde(mu)^^ tilde(nu)=(-1)^(pq) tilde(nu)^^ tilde(mu):}\begin{equation*} \tilde{\boldsymbol{\mu}} \wedge \tilde{\boldsymbol{\nu}}=(-1)^{p q} \tilde{\boldsymbol{\nu}} \wedge \tilde{\boldsymbol{\mu}} \tag{32.12} \end{equation*}(32.12)μ~ν~=(1)pqν~μ~
Using this idea, we can generate p p ppp-forms by making more and more wedge products. 6 6 ^(6){ }^{6}6
A p-form α ~ α ~ tilde(alpha)\tilde{\boldsymbol{\alpha}}α~ can be written as
α ~ = 1 p ! α i 1 , i 2 , , i p ω i 1 ω i 2 ω i p (32.13) = α | i 1 , i 2 , , i p | ω i 1 ω i 2 ω i p α ~ = 1 p ! α i 1 , i 2 , , i p ω i 1 ω i 2 ω i p (32.13) = α i 1 , i 2 , , i p ω i 1 ω i 2 ω i p {:[ tilde(alpha)=(1)/(p!)alpha_(i_(1),i_(2),dots,i_(p))omega^(i_(1))^^omega^(i_(2))^^dots^^omega^(i_(p))],[(32.13)=alpha_(|i_(1),i_(2),dots,i_(p)|)omega^(i_(1))^^omega^(i_(2))^^dots^^omega^(i_(p))]:}\begin{align*} \tilde{\boldsymbol{\alpha}} & =\frac{1}{p!} \alpha_{i_{1}, i_{2}, \ldots, i_{p}} \boldsymbol{\omega}^{i_{1}} \wedge \boldsymbol{\omega}^{i_{2}} \wedge \ldots \wedge \boldsymbol{\omega}^{i_{p}} \\ & =\alpha_{\left|i_{1}, i_{2}, \ldots, i_{p}\right|} \boldsymbol{\omega}^{i_{1}} \wedge \boldsymbol{\omega}^{i_{2}} \wedge \ldots \wedge \boldsymbol{\omega}^{i_{p}} \tag{32.13} \end{align*}α~=1p!αi1,i2,,ipωi1ωi2ωip(32.13)=α|i1,i2,,ip|ωi1ωi2ωip
Here, the vertical bars around a list of i n i n i_(n)i_{n}in mean we fix i 1 < i 2 < < i p i 1 < i 2 < < i p i_(1) < i_(2) < dots < i_(p)i_{1}<i_{2}<\ldots<i_{p}i1<i2<<ip and don't therefore need the 1 / p 1 / p 1//p1 / p1/p ! normalization in the first. 7 7 ^(7){ }^{7}7
Example 32.4
Let's make a 3 -form from the 1 -forms σ ~ , τ ~ σ ~ , τ ~ tilde(sigma), tilde(tau)\tilde{\boldsymbol{\sigma}}, \tilde{\boldsymbol{\tau}}σ~,τ~ and κ ~ κ ~ tilde(kappa)\tilde{\boldsymbol{\kappa}}κ~. We have
σ ~ ( τ ~ κ ~ ) = ( σ ~ τ ~ ) κ ~ = σ ~ τ ~ κ ~ σ ~ κ ~ τ ~ + κ ~ σ ~ τ ~ κ ~ τ ~ σ ~ (32.15) + τ ~ κ ~ σ ~ τ ~ σ ~ κ ~ . σ ~ ( τ ~ κ ~ ) = ( σ ~ τ ~ ) κ ~ = σ ~ τ ~ κ ~ σ ~ κ ~ τ ~ + κ ~ σ ~ τ ~ κ ~ τ ~ σ ~ (32.15) + τ ~ κ ~ σ ~ τ ~ σ ~ κ ~ . {:[ tilde(sigma)^^( tilde(tau)^^ tilde(kappa))=( tilde(sigma)^^ tilde(tau))^^ tilde(kappa)= tilde(sigma)ox tilde(tau)ox tilde(kappa)- tilde(sigma)ox tilde(kappa)ox tilde(tau)],[+ tilde(kappa)ox tilde(sigma)ox tilde(tau)- tilde(kappa)ox tilde(tau)ox tilde(sigma)],[(32.15)+ tilde(tau)ox tilde(kappa)ox tilde(sigma)- tilde(tau)ox tilde(sigma)ox tilde(kappa).]:}\begin{align*} \tilde{\boldsymbol{\sigma}} \wedge(\tilde{\boldsymbol{\tau}} \wedge \tilde{\boldsymbol{\kappa}})=(\tilde{\boldsymbol{\sigma}} \wedge \tilde{\boldsymbol{\tau}}) \wedge \tilde{\boldsymbol{\kappa}}= & \tilde{\boldsymbol{\sigma}} \otimes \tilde{\boldsymbol{\tau}} \otimes \tilde{\boldsymbol{\kappa}}-\tilde{\boldsymbol{\sigma}} \otimes \tilde{\boldsymbol{\kappa}} \otimes \tilde{\boldsymbol{\tau}} \\ & +\tilde{\boldsymbol{\kappa}} \otimes \tilde{\boldsymbol{\sigma}} \otimes \tilde{\boldsymbol{\tau}}-\tilde{\boldsymbol{\kappa}} \otimes \tilde{\boldsymbol{\tau}} \otimes \tilde{\boldsymbol{\sigma}} \\ & +\tilde{\boldsymbol{\tau}} \otimes \tilde{\boldsymbol{\kappa}} \otimes \tilde{\boldsymbol{\sigma}}-\tilde{\boldsymbol{\tau}} \otimes \tilde{\boldsymbol{\sigma}} \otimes \tilde{\boldsymbol{\kappa}} . \tag{32.15} \end{align*}σ~(τ~κ~)=(σ~τ~)κ~=σ~τ~κ~σ~κ~τ~+κ~σ~τ~κ~τ~σ~(32.15)+τ~κ~σ~τ~σ~κ~.
Geometrically, we can build this object by stepping up from the equally spaced planes of the 1-form, via the tube-like structure of the 2 -form to the cellular structure of the 3 -form.
A simple example of a p p ppp-form that we can build in n n nnn-dimensional space involves multiplying a scalar function α | i 1 , i 2 i p | = f ( x 1 , x 2 , , x p ) α i 1 , i 2 i p = f x 1 , x 2 , , x p alpha_(|i_(1),i_(2)dotsi_(p)|)=f(x^(1),x^(2),dots,x^(p))\alpha_{\left|i_{1}, i_{2} \ldots i_{p}\right|}=f\left(x^{1}, x^{2}, \ldots, x^{p}\right)α|i1,i2ip|=f(x1,x2,,xp) by wedge products of the basis 1 -forms or, equivalently, the differentials of coordinates to build a p p ppp-form tensor β ~ β ~ tilde(beta)\tilde{\boldsymbol{\beta}}β~
(32.16) β ~ = f ( x 1 , x 2 , , x p ) d x i 1 d x i 2 d x i p (32.16) β ~ = f x 1 , x 2 , , x p d x i 1 d x i 2 d x i p {:(32.16) tilde(beta)=f(x^(1),x^(2),dots,x^(p))dx^(i_(1))^^dx^(i_(2))^^dots^^dx^(i_(p)):}\begin{equation*} \tilde{\boldsymbol{\beta}}=f\left(x^{1}, x^{2}, \ldots, x^{p}\right) \boldsymbol{d} x^{i_{1}} \wedge \boldsymbol{d} x^{i_{2}} \wedge \ldots \wedge \boldsymbol{d} x^{i_{p}} \tag{32.16} \end{equation*}(32.16)β~=f(x1,x2,,xp)dxi1dxi2dxip
All p p ppp-forms are linear and can therefore be added to other p p ppp-forms to make the arbitrary p p ppp-form of your choosing.

Example 32.5

In three-dimensional space with basis 1 -forms ω 1 = d x , ω 2 = d y ω 1 = d x , ω 2 = d y omega^(1)=dx,omega^(2)=dy\boldsymbol{\omega}^{1}=\boldsymbol{d} x, \boldsymbol{\omega}^{2}=\boldsymbol{d} yω1=dx,ω2=dy and ω 3 = d z ω 3 = d z omega^(3)=dz\boldsymbol{\omega}^{3}=\boldsymbol{d} zω3=dz, we can investigate some p p ppp-form objects that we make from functions of the coordinates ( x , y , z ) ( x , y , z ) (x,y,z)(x, y, z)(x,y,z) and the three basis forms:
0 -form α ~ = f ( x , y , z ) α ~ = f ( x , y , z ) quad tilde(alpha)=f(x,y,z)\quad \tilde{\boldsymbol{\alpha}}=f(x, y, z)α~=f(x,y,z),
1-form β ~ = A ( x , y , z ) d x + B ( x , y , z ) d y + C ( x , y , z ) d z β ~ = A ( x , y , z ) d x + B ( x , y , z ) d y + C ( x , y , z ) d z quad tilde(beta)=A(x,y,z)dx+B(x,y,z)dy+C(x,y,z)dz\quad \tilde{\boldsymbol{\beta}}=A(x, y, z) \boldsymbol{d} x+B(x, y, z) \boldsymbol{d} y+C(x, y, z) \boldsymbol{d} zβ~=A(x,y,z)dx+B(x,y,z)dy+C(x,y,z)dz,
2-form μ ~ = D ( x , y , z ) d x d y + E ( x , y , z ) d y d z + F ( x , y , z ) d z d x μ ~ = D ( x , y , z ) d x d y + E ( x , y , z ) d y d z + F ( x , y , z ) d z d x quad tilde(mu)=D(x,y,z)dx^^dy+E(x,y,z)dy^^dz+F(x,y,z)dz^^dx\quad \tilde{\boldsymbol{\mu}}=D(x, y, z) \boldsymbol{d} x \wedge \boldsymbol{d} y+E(x, y, z) \boldsymbol{d} y \wedge \boldsymbol{d} z+F(x, y, z) \boldsymbol{d} z \wedge \boldsymbol{d} xμ~=D(x,y,z)dxdy+E(x,y,z)dydz+F(x,y,z)dzdx, 3-form ν ~ = G ( x , y , z ) d x d y d z ν ~ = G ( x , y , z ) d x d y d z quad tilde(nu)=G(x,y,z)dx^^dy^^dz\quad \tilde{\nu}=G(x, y, z) \boldsymbol{d} x \wedge \boldsymbol{d} y \wedge \boldsymbol{d} zν~=G(x,y,z)dxdydz.
Notice how having two identical 1-forms in any wedge product causes it to vanish (e.g. d x d y d = 0 d x d y d = 0 dx^^dy^^d=0\boldsymbol{d} x \wedge \boldsymbol{d} y \wedge \boldsymbol{d}=0dxdyd=0 ). This constrains the possible p p ppp-forms to those above. It is not possible, for example, for a 4 -form to exist in three spatial dimensions.

32.3 p 32.3 p 32.3 p32.3 p32.3p-vectors

Consider vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v in Euclidean 2-space. We can build a parallelogram in Fig. 32.4 from these vectors. Do this by starting at the origin and drawing u u u\boldsymbol{u}u and then v v v\boldsymbol{v}v, returning to the origin and drawing v v v\boldsymbol{v}v and then u u u\boldsymbol{u}u. The area of this parallelogram is given by the computation of the determinant
(32.18) (Area) = | u x u y v x v y | = u x v y v x u y = ε i j u i v j (32.18)  (Area)  = u x u y v x v y = u x v y v x u y = ε i j u i v j {:(32.18)" (Area) "=|[u^(x),u^(y)],[v^(x),v^(y)]|=u^(x)v^(y)-v^(x)u^(y)=epsi_(ij)u^(i)v^(j):}\text { (Area) }=\left|\begin{array}{cc} u^{x} & u^{y} \tag{32.18}\\ v^{x} & v^{y} \end{array}\right|=u^{x} v^{y}-v^{x} u^{y}=\varepsilon_{i j} u^{i} v^{j}(32.18) (Area) =|uxuyvxvy|=uxvyvxuy=εijuivj
where ε i j ε i j epsi_(ij)\varepsilon_{i j}εij is the two-dimensional Levi-Civita symbol. 8 8 ^(8){ }^{8}8 The signed area is antisymmetric in the components of u u u\boldsymbol{u}u and v v v\boldsymbol{v}v, which can be thought of as giving a handedness to the area. The area of the parallelogram is related to the bivector b b b\boldsymbol{b}b defined as the antisymmetric ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor 9 9 ^(9)^{9}9
(32.19) b ( , ) = u v = u v v u (32.19) b ( , ) = u v = u v v u {:(32.19)b(",")=u^^v=u ox v-v ox u:}\begin{equation*} \boldsymbol{b}(,)=\boldsymbol{u} \wedge \boldsymbol{v}=\boldsymbol{u} \otimes \boldsymbol{v}-\boldsymbol{v} \otimes \boldsymbol{u} \tag{32.19} \end{equation*}(32.19)b(,)=uv=uvvu
In fact, it is fairly easy to see that the area is of the parallelogram is equal to the component b x y b x y b^(xy)b^{x y}bxy, accessible by inserting basis 1 -forms ω x ω x omega^(x)\boldsymbol{\omega}^{x}ωx and ω y ω y omega^(y)\boldsymbol{\omega}^{y}ωy into the slots of the bivector [i.e. b ( ω x , ω y ) b ω x , ω y b(omega^(x),omega^(y))\boldsymbol{b}\left(\boldsymbol{\omega}^{x}, \boldsymbol{\omega}^{y}\right)b(ωx,ωy) ]. We will return to this in Chapter 37.
For now, we note that the existence of p p ppp-vectors, formed from the wedge products of vectors with components
v = 1 p ! v i 1 , i 2 i p e i 1 e i 2 e i p (32.20) = v | i 1 , i 2 i p | e i 1 e i 2 e i p v = 1 p ! v i 1 , i 2 i p e i 1 e i 2 e i p (32.20) = v i 1 , i 2 i p e i 1 e i 2 e i p {:[v=(1)/(p!)v^(i_(1),i_(2)dotsi_(p))e_(i_(1))^^e_(i_(2))^^dots^^e_(i_(p))],[(32.20)=v^(|i_(1),i_(2)dotsi_(p)|)e_(i_(1))^^e_(i_(2))^^dots^^e_(i_(p))]:}\begin{align*} \boldsymbol{v}= & \frac{1}{p!} v^{i_{1}, i_{2} \ldots i_{p}} \boldsymbol{e}_{i_{1}} \wedge \boldsymbol{e}_{i_{2}} \wedge \ldots \wedge \boldsymbol{e}_{i_{p}} \\ & =v^{\left|i_{1}, i_{2} \ldots i_{p}\right|} \boldsymbol{e}_{i_{1}} \wedge \boldsymbol{e}_{i_{2}} \wedge \ldots \wedge \boldsymbol{e}_{i_{p}} \tag{32.20} \end{align*}v=1p!vi1,i2ipei1ei2eip(32.20)=v|i1,i2ip|ei1ei2eip
8 8 ^(8){ }^{8}8 In two dimensions, ε 12 = 1 , ε 21 = ε 12 = 1 , ε 21 = epsi_(12)=1,epsi_(21)=\varepsilon_{12}=1, \varepsilon_{21}=ε12=1,ε21= -1 , and ε 11 = ε 22 = 0 ε 11 = ε 22 = 0 epsi_(11)=epsi_(22)=0\varepsilon_{11}=\varepsilon_{22}=0ε11=ε22=0. In general, ε i 1 i n = ( 1 ) P ε i 1 i n = ( 1 ) P epsi_(i_(1)dotsi_(n))=(-1)^(P)\varepsilon_{i_{1} \ldots i_{n}}=(-1)^{P}εi1in=(1)P, where P P PPP is the parity of the permutation of 1 , 2 , 3 n 1 , 2 , 3 n 1,2,3dots n1,2,3 \ldots n1,2,3n in the indices (i.e. how many pairwise swaps need to be made to a string of numbers to work it back to the order 1 n 1 n 1dots n1 \ldots n1n ), and ε i 1 i n = 0 ε i 1 i n = 0 epsi_(i_(1)dotsi_(n))=0\varepsilon_{i_{1} \ldots i_{n}}=0εi1in=0 if any of the indices are repeated.
Fig. 32.4 The parallelogram formed by vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v.
9 9 ^(9){ }^{9}9 As for the case of forms, we can write the components of a bivector as
b = 1 2 b μ ν e μ e ν b = 1 2 b μ ν e μ e ν b=(1)/(2)b^(mu nu)e_(mu)^^e_(nu)\boldsymbol{b}=\frac{1}{2} b^{\mu \nu} \boldsymbol{e}_{\mu} \wedge \boldsymbol{e}_{\nu}b=12bμνeμeν
10 10 ^(10){ }^{10}10 The symbol δ j 1 j p i 1 i p δ j 1 j p i 1 i p delta_(j_(1)dotsj_(p))^(i_(1)dotsi_(p))\delta_{j_{1} \ldots j_{p}}^{i_{1} \ldots i_{p}}δj1jpi1ip is defined as
δ j 1 j p i 1 i p = δ j 1 j p i 1 i p = delta_(j_(1)dotsj_(p))^(i_(1)dotsi_(p))=\delta_{j_{1} \ldots j_{p}}^{i_{1} \ldots i_{p}}=δj1jpi1ip=
{ + 1 if ( i 1 i p ) is an even permutation of ( j 1 j p ) 1 if ( i 1 i p ) is an odd permutation of ( j 1 j p ) 0 any two i s are the same 0 any two j s are the same 0 the is and j s are different integers. + 1  if  i 1 i p  is an even   permutation of  j 1 j p 1  if  i 1 i p  is an odd   permutation of  j 1 j p 0  any two  i  s are the same  0  any two  j  s are the same  0  the is and  j  s are   different integers.  {[+1," if "(i_(1)dotsi_(p))" is an even "],[," permutation of "(j_(1)dotsj_(p))],[-1," if "(i_(1)dotsi_(p))" is an odd "],[," permutation of "(j_(1)dotsj_(p))],[0," any two "i" s are the same "],[0," any two "j" s are the same "],[0," the is and "j" s are "],[," different integers. "]:}\begin{cases}+1 & \text { if }\left(i_{1} \ldots i_{p}\right) \text { is an even } \\ & \text { permutation of }\left(j_{1} \ldots j_{p}\right) \\ -1 & \text { if }\left(i_{1} \ldots i_{p}\right) \text { is an odd } \\ & \text { permutation of }\left(j_{1} \ldots j_{p}\right) \\ 0 & \text { any two } i \text { s are the same } \\ 0 & \text { any two } j \text { s are the same } \\ 0 & \text { the is and } j \text { s are } \\ & \text { different integers. }\end{cases}{+1 if (i1ip) is an even  permutation of (j1jp)1 if (i1ip) is an odd  permutation of (j1jp)0 any two i s are the same 0 any two j s are the same 0 the is and j s are  different integers. 
Fig. 32.5 The bivector b b b\boldsymbol{b}b and 2-form F ~ F ~ tilde(F)\tilde{\boldsymbol{F}}F~ are used to for the inner product F ~ , b F ~ , b (: tilde(F),b:)\langle\tilde{\boldsymbol{F}}, \boldsymbol{b}\rangleF~,b. This outputs a number equal to the number of tubes of F ~ F ~ tilde(F)\tilde{\boldsymbol{F}}F~ contained within the parallelogram b b b\boldsymbol{b}b, which in this case is 4 .
The existence of p p ppp-vectors allows us to define a set of generalized inner products of p p ppp-forms, such that p p ppp-vectors and p p ppp-forms can be mapped on to numbers. In order to make the inner product of a p p ppp-form α ~ = α ~ = tilde(alpha)=\tilde{\boldsymbol{\alpha}}=α~= α | i 1 i p | ω i 1 ω i p α i 1 i p ω i 1 ω i p alpha_(|i_(1)dotsi_(p)|)omega^(i_(1))^^dots^^omega^(i_(p))\alpha_{\left|i_{1} \ldots i_{p}\right|} \boldsymbol{\omega}^{i_{1}} \wedge \ldots \wedge \boldsymbol{\omega}^{i_{p}}α|i1ip|ωi1ωip and a p p ppp-vector v = v | j 1 j p | e j 1 e j p v = v j 1 j p e j 1 e j p v=v^(|j_(1)dotsj_(p)|)e_(j_(1))^^dots^^e_(j_(p))\boldsymbol{v}=v^{\left|j_{1} \ldots j_{p}\right|} \boldsymbol{e}_{j_{1}} \wedge \ldots \wedge \boldsymbol{e}_{j_{p}}v=v|j1jp|ej1ejp, we have the rule 10 10 ^(10){ }^{10}10
α ~ , v = α | i 1 i p | v | j 1 j p | ω i 1 ω i p , e j 1 e j p = α | i 1 i p | v | j 1 j p | δ j 1 j p i 1 i p (32.21) = α | i 1 i p | v i 1 i p α ~ , v = α i 1 i p v j 1 j p ω i 1 ω i p , e j 1 e j p = α i 1 i p v j 1 j p δ j 1 j p i 1 i p (32.21) = α i 1 i p v i 1 i p {:[(: tilde(alpha)","v:)=alpha_(|i_(1)dotsi_(p)|)v^(|j_(1)dotsj_(p)|)(:omega^(i_(1))^^dots^^omega^(i_(p)),e_(j_(1))^^dots^^e_(j_(p)):)],[=alpha_(|i_(1)dotsi_(p)|)v^(|j_(1)dotsj_(p)|)delta_(j_(1)dotsj_(p))^(i_(1)dotsi_(p))],[(32.21)=alpha_(|i_(1)dotsi_(p)|)v^(i_(1)dotsi_(p))]:}\begin{align*} \langle\tilde{\boldsymbol{\alpha}}, \boldsymbol{v}\rangle & =\alpha_{\left|i_{1} \ldots i_{p}\right|} v^{\left|j_{1} \ldots j_{p}\right|}\left\langle\boldsymbol{\omega}^{i_{1}} \wedge \ldots \wedge \boldsymbol{\omega}^{i_{p}}, \boldsymbol{e}_{j_{1}} \wedge \ldots \wedge \boldsymbol{e}_{j_{p}}\right\rangle \\ & =\alpha_{\left|i_{1} \ldots i_{p}\right|} v^{\left|j_{1} \ldots j_{p}\right|} \delta_{j_{1} \ldots j_{p}}^{i_{1} \ldots i_{p}} \\ & =\alpha_{\left|i_{1} \ldots i_{p}\right|} v^{i_{1} \ldots i_{p}} \tag{32.21} \end{align*}α~,v=α|i1ip|v|j1jp|ωi1ωip,ej1ejp=α|i1ip|v|j1jp|δj1jpi1ip(32.21)=α|i1ip|vi1ip
Example 32.6
Work in three spatial dimensions, with components arranged in the order ( x 1 , x 2 , x 3 ) = ( x , y , z ) x 1 , x 2 , x 3 = ( x , y , z ) (x^(1),x^(2),x^(3))=(x,y,z)\left(x^{1}, x^{2}, x^{3}\right)=(x, y, z)(x1,x2,x3)=(x,y,z). Combine a 2 -form F ~ F ~ tilde(F)\tilde{\boldsymbol{F}}F~ and a bivector b b b\boldsymbol{b}b to make a number thus
(32.22) F ~ , b = F x y b x y + F x z b x z + F y z b y z (32.22) F ~ , b = F x y b x y + F x z b x z + F y z b y z {:(32.22)(: tilde(F)","b:)=F_(xy)b^(xy)+F_(xz)b^(xz)+F_(yz)b^(yz):}\begin{equation*} \langle\tilde{\boldsymbol{F}}, \boldsymbol{b}\rangle=F_{x y} b^{x y}+F_{x z} b^{x z}+F_{y z} b^{y z} \tag{32.22} \end{equation*}(32.22)F~,b=Fxybxy+Fxzbxz+Fyzbyz
This has a pleasing resemblance to the dot product of vectors, or the inner product of a vector and a 1 -form. The geometrical interpretation of this is that the scalar outputted by the inner product represents the number of tubes of the 2 -form F ~ F ~ tilde(F)\tilde{\boldsymbol{F}}F~ contained in the parallelogram representing the bivector b b b\boldsymbol{b}b (Fig. 32.5).
It's also useful to note that if the bivector b b b\boldsymbol{b}b is formed from vectors c c c\boldsymbol{c}c and d d d\boldsymbol{d}d via b = c d b = c d b=c^^d\boldsymbol{b}=\boldsymbol{c} \wedge \boldsymbol{d}b=cd then we also have that
(32.23) F ~ , c d = F ~ ( c , d ) (32.23) F ~ , c d = F ~ ( c , d ) {:(32.23)(: tilde(F)","c^^d:)= tilde(F)(c","d):}\begin{equation*} \langle\tilde{\boldsymbol{F}}, \boldsymbol{c} \wedge \boldsymbol{d}\rangle=\tilde{\boldsymbol{F}}(\boldsymbol{c}, \boldsymbol{d}) \tag{32.23} \end{equation*}(32.23)F~,cd=F~(c,d)
This is proved in the exercises.
To add to our list of known tensor objects, we now have p p ppp-forms and q q qqq-vectors, which are examples of antisymmetric tensors. In the next chapter, our study of differential geometry starts in earnest, as we start to take derivatives.

Chapter summary

  • Differential forms can be built from 1-forms using the wedge product. A p p ppp-form is an antisymmetric ( 0 , p ) ( 0 , p ) (0,p)(0, p)(0,p) tensor.
  • 2 -forms can be represented as a tube-like structure with a handedness.
  • p p ppp-vectors can be built from vectors using the wedge product. A p p ppp-vector and a p p ppp-form can be combined to make a number.

Exercises

(32.1) Verify eqn 32.23 by computing components.
(32.2) Fermi transport: Consider a sphere, and vectors a , b a , b a,b\boldsymbol{a}, \boldsymbol{b}a,b and c c c\boldsymbol{c}c that are orthogonal to each other and to the velocity vector u u u\boldsymbol{u}u of the sphere (i.e. the vector field that is tangent to the sphere's world line). The three vectors determine the orientation of the sphere along its world line and can be used to measure the rotation of the sphere. We can write down the equations of motion that tells us there is no rotation. The extent to which this equation is satisfied measures the rotation. The equations of motion, known as the Fermi transport equations,
are
u a = ( A u ) ( , a ~ ) u b = ( A u ) ( , b ~ ) (32.24) u c = ( A u ) ( , c ~ ) u a = ( A u ) ( , a ~ ) u b = ( A u ) ( , b ~ ) (32.24) u c = ( A u ) ( , c ~ ) {:[grad_(u)a=-(A^^u)("," tilde(a))],[grad_(u)b=-(A^^u)("," tilde(b))],[(32.24)grad_(u)c=-(A^^u)("," tilde(c))]:}\begin{align*} \boldsymbol{\nabla}_{u} \boldsymbol{a} & =-(\boldsymbol{A} \wedge \boldsymbol{u})(, \tilde{a}) \\ \boldsymbol{\nabla}_{u} \boldsymbol{b} & =-(\boldsymbol{A} \wedge \boldsymbol{u})(, \tilde{b}) \\ \boldsymbol{\nabla}_{u} \boldsymbol{c} & =-(\boldsymbol{A} \wedge \boldsymbol{u})(, \tilde{\boldsymbol{c}}) \tag{32.24} \end{align*}ua=(Au)(,a~)ub=(Au)(,b~)(32.24)uc=(Au)(,c~)
where A = u u A = u u A=grad_(u)u\boldsymbol{A}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}A=uu is the acceleration.
(a) Write one of these equations in components.
(b) What happens for transport along a geodesic world line?
(c) Show that if a , b a , b a,b\boldsymbol{a}, \boldsymbol{b}a,b and c c c\boldsymbol{c}c are initially orthogonal, they stay orthogonal.
(d) Show that a vector transported by these equations remains orthogonal to the four-velocity.
(e) Compare Fermi transport with parallel transport.

33

33.1 Exterior calculus 340 33.2 Commutators 342 33.3 Lie derivatives of vectors 344 33.4 Lie derivatives of tensors 347 33.5 Killing vectors 348 Chapter summary 350 Exercises 350
Fig. 33.1 The action of the exterior derivative d d d\boldsymbol{d}d, on the 1 -form f ( x ) d y f ( x ) d y f(x)dyf(x) \boldsymbol{d} yf(x)dy is to convert it into f ( x ) / x ( d x d y ) f ( x ) / x ( d x d y ) del f(x)//del x(dx^^dy)\partial f(x) / \partial x(\boldsymbol{d} x \wedge \boldsymbol{d} y)f(x)/x(dxdy).

Exterior and Lie derivatives

More ways of killing a cat than choking her with cream

Charles Kingsley (1819-1875)
Derivatives are undeniably important in physics. We would like a way of taking derivatives of tensor fields that does not rely on coordinates. We naturally reach the idea of a derivative if we think in terms of a tensors changing as we move around a space. More precisely, we work with the notion of a tensor field which, at each point in space, provides us with a tensor. In working out derivatives, we are faced with the problem that there is no obvious way to compare tensor fields evaluated at two different points in a space. One way around this is to use the metric to define a notion of parallelism. This requires a connection: a method of connecting different points, so that tensors can be moved in a parallel manner around the space. We examine this connection in more detail in Chapter 34. If we don't have this parallelism (or we don't have a metric) then how can we proceed? In this chapter, we examine two ways of dealing with more primitive notions of rates of change of tensors. The first type of derivative applies only to fields of forms and is called the exterior derivative. We shall later use this derivative to describe the curvature of spacetime. The second type of derivative applies to all tensors and relies on being able to carry the tensor field of interest across the space using a separate vector field. This is called a Lie derivative and is the subject of the second part of this chapter. The Lie derivative is especially useful in cosmology for describing changes as we travel along with the cosmological fluid. However, perhaps its most useful feature is in identifying conserved quantities via Killing vectors, which we describe at the end of the chapter.

33.1 Exterior calculus

The exterior derivative is a differential operation that can be applied to forms. In measuring a rate of change of a p p ppp-form, exterior differentiation converts the p p ppp-form into a ( p + 1 ) ( p + 1 ) (p+1)(p+1)(p+1)-form. This has a pictorial interpretation which is clearest for the action of the exterior derivative on a 1 -form and is shown in Fig. 33.1. Recall that a 1 -form can be thought of as a set of parallel surfaces in space, while a 2 -form is a tubular structure. By measuring how fast a component of the 1 -form changes along a some direction, the exterior derivative builds us a 2 -form by adding surfaces normal to this direction.
To use the exterior derivative we define an operator d d d\boldsymbol{d}d that acts on forms. The simplest 1 -form is the differential d f d f df\boldsymbol{d} fdf. This 1 -form can be thought of as originating from the action of the exterior derivative operator d d d\boldsymbol{d}d on the 0 -form f ( x ) f ( x ) f(x)f(x)f(x). In general, the operator d d ~ d d ~ d_( tilde(d))\underset{\tilde{d}}{\boldsymbol{d}}dd~ can be thought of as a ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1) object that creates the ( p + 1 ) ( p + 1 ) (p+1)(p+1)(p+1)-form d A ~ d A ~ d tilde(A)\boldsymbol{d} \tilde{\boldsymbol{A}}dA~ from a p p ppp-form A ~ A ~ tilde(A)\tilde{\boldsymbol{A}}A~. It obeys the rules listed in the margin.
The rules feel rather austere. We get a clearer idea of what the operator does in practice if we consider its action on an arbitrary p p ppp-form
(33.6) A ~ = 1 p ! A α , β σ d x α d x β d x σ (33.6) A ~ = 1 p ! A α , β σ d x α d x β d x σ {:(33.6) tilde(A)=(1)/(p!)A_(alpha,beta dots sigma)dx^(alpha)^^dx^(beta)^^dots^^dx^(sigma):}\begin{equation*} \tilde{\boldsymbol{A}}=\frac{1}{p!} A_{\alpha, \beta \ldots \sigma} \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \wedge \ldots \wedge \boldsymbol{d} x^{\sigma} \tag{33.6} \end{equation*}(33.6)A~=1p!Aα,βσdxαdxβdxσ
To be consistent with the rules, we must have
(33.7) d A ~ = 1 p ! d A α β σ d x α d x β d x σ (33.7) d A ~ = 1 p ! d A α β σ d x α d x β d x σ {:(33.7)d tilde(A)=(1)/(p!)dA_(alpha beta dots sigma)^^dx^(alpha)^^dx^(beta)^^dots^^dx^(sigma):}\begin{equation*} \boldsymbol{d} \tilde{\boldsymbol{A}}=\frac{1}{p!} \boldsymbol{d} A_{\alpha \beta \ldots \sigma} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \wedge \ldots \wedge \boldsymbol{d} x^{\sigma} \tag{33.7} \end{equation*}(33.7)dA~=1p!dAαβσdxαdxβdxσ
where
(33.8) d A α β σ = A α β σ x μ d x μ (33.8) d A α β σ = A α β σ x μ d x μ {:(33.8)dA_(alpha beta dots sigma)=(delA_(alpha beta dots sigma))/(delx^(mu))dx^(mu):}\begin{equation*} \boldsymbol{d} A_{\alpha \beta \ldots \sigma}=\frac{\partial A_{\alpha \beta \ldots \sigma}}{\partial x^{\mu}} \boldsymbol{d} x^{\mu} \tag{33.8} \end{equation*}(33.8)dAαβσ=Aαβσxμdxμ
As a result, we have the following: 1 1 ^(1){ }^{1}1
The exterior derivative of a p p ppp-form α α alpha\boldsymbol{\alpha}α is given by
(33.10) d A ~ = 1 p ! A α β σ x μ d x μ d x α d x β d x σ (33.10) d A ~ = 1 p ! A α β σ x μ d x μ d x α d x β d x σ {:(33.10)d tilde(A)=(1)/(p!)(delA_(alpha beta dots sigma))/(delx^(mu))dx^(mu)^^dx^(alpha)^^dx^(beta)^^dots^^dx^(sigma):}\begin{equation*} \boldsymbol{d} \tilde{\boldsymbol{A}}=\frac{1}{p!} \frac{\partial A_{\alpha \beta \ldots \sigma}}{\partial x^{\mu}} \boldsymbol{d} x^{\mu} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \wedge \ldots \wedge \boldsymbol{d} x^{\sigma} \tag{33.10} \end{equation*}(33.10)dA~=1p!Aαβσxμdxμdxαdxβdxσ
As with all calculus, the only way to really see what's going on is to examine some examples.
Example 33.1
We work in (3+1)-dimensional spacetime with spherical coordinates x μ = ( t , χ , θ , ϕ ) x μ = ( t , χ , θ , ϕ ) x^(mu)=(t,chi,theta,phi)x^{\mu}=(t, \chi, \theta, \phi)xμ=(t,χ,θ,ϕ).
We act with d d d\boldsymbol{d}d on a 1 -form A ~ = A α ( t , χ , θ , ϕ ) d x α A ~ = A α ( t , χ , θ , ϕ ) d x α tilde(A)=A_(alpha)(t,chi,theta,phi)dx^(alpha)\tilde{\boldsymbol{A}}=A_{\alpha}(t, \chi, \theta, \phi) \boldsymbol{d} x^{\alpha}A~=Aα(t,χ,θ,ϕ)dxα, where A α A α A_(alpha)A_{\alpha}Aα are functions of the coordinates. We will obtain
(33.11) d A ~ = A α x μ d x μ d x α (33.11) d A ~ = A α x μ d x μ d x α {:(33.11)d tilde(A)=(delA_(alpha))/(delx^(mu))dx^(mu)^^dx^(alpha):}\begin{equation*} \boldsymbol{d} \tilde{\boldsymbol{A}}=\frac{\partial A_{\alpha}}{\partial x^{\mu}} \boldsymbol{d} x^{\mu} \wedge \boldsymbol{d} x^{\alpha} \tag{33.11} \end{equation*}(33.11)dA~=Aαxμdxμdxα
Example 0: Consider a 1 -form A ~ = d A ~ = d tilde(A)=d\tilde{\boldsymbol{A}}=\boldsymbol{d}A~=d. Acting with d d d\boldsymbol{d}d we obtain
(33.12) d A ~ = d d t = 0 (33.12) d A ~ = d d t = 0 {:(33.12)d tilde(A)=ddt=0:}\begin{equation*} d \tilde{A}=d d t=0 \tag{33.12} \end{equation*}(33.12)dA~=ddt=0
where we've used d d = 0 d d = 0 dd=0\boldsymbol{d} \boldsymbol{d}=0dd=0 (rule IV in the box).
Example 1: Consider a 1-form A ~ = a ( t ) d χ A ~ = a ( t ) d χ tilde(A)=a(t)d chi\tilde{\boldsymbol{A}}=a(t) \boldsymbol{d} \chiA~=a(t)dχ, where a ( t ) a ( t ) a(t)a(t)a(t) is a scalar function of t t ttt only.
Acting with d d d\boldsymbol{d}d we obtain
(33.13) d A ~ = a ( t ) t d t d χ (33.13) d A ~ = a ( t ) t d t d χ {:(33.13)d tilde(A)=(del a(t))/(del t)*dt^^d chi:}\begin{equation*} \boldsymbol{d} \tilde{\boldsymbol{A}}=\frac{\partial a(t)}{\partial t} \cdot \boldsymbol{d} t \wedge \boldsymbol{d} \chi \tag{33.13} \end{equation*}(33.13)dA~=a(t)tdtdχ
where we've used the fact that a ( t ) / x i = 0 a ( t ) / x i = 0 del a(t)//delx^(i)=0\partial a(t) / \partial x^{i}=0a(t)/xi=0, for i t i t i!=ti \neq tit.
Example 2: Consider a 1 -form A = a ( t ) sin θ d θ A = a ( t ) sin θ d θ vec(A)=a(t)sin theta d theta\overrightarrow{\boldsymbol{A}}=a(t) \sin \theta \boldsymbol{d} \thetaA=a(t)sinθdθ. Taking the exterior derivative we obtain
d A ~ = a ( t ) t sin θ d t d θ + a ( t ) cos θ d θ d θ (33.14) = a ( t ) t sin θ d t d θ . d A ~ = a ( t ) t sin θ d t d θ + a ( t ) cos θ d θ d θ (33.14) = a ( t ) t sin θ d t d θ . {:[d tilde(A)=(del a(t))/(del t)sin theta dt^^d theta+a(t)cos theta d theta^^d theta],[(33.14)=(del a(t))/(del t)sin theta dt^^d theta.]:}\begin{align*} \boldsymbol{d} \tilde{\boldsymbol{A}} & =\frac{\partial a(t)}{\partial t} \sin \theta \boldsymbol{d} t \wedge \boldsymbol{d} \theta+a(t) \cos \theta \boldsymbol{d} \theta \wedge \boldsymbol{d} \theta \\ & =\frac{\partial a(t)}{\partial t} \sin \theta \boldsymbol{d} t \wedge \boldsymbol{d} \theta . \tag{33.14} \end{align*}dA~=a(t)tsinθdtdθ+a(t)cosθdθdθ(33.14)=a(t)tsinθdtdθ.
where we've used the fact that d x μ d x μ = 0 d x μ d x μ = 0 dx^(mu)^^dx^(mu)=0\boldsymbol{d} x^{\mu} \wedge \boldsymbol{d} x^{\mu}=0dxμdxμ=0.
The exterior derivative operator d d d\boldsymbol{d}d has the following properties when acting on the p p ppp-form α ~ α ~ tilde(alpha)\tilde{\boldsymbol{\alpha}}α~ and q q qqq-form β β beta\boldsymbol{\beta}β : I: It is linear
(33.1) d ( α ~ + β ~ ) = d α ~ + d β ~ (33.1) d ( α ~ + β ~ ) = d α ~ + d β ~ {:(33.1)d( tilde(alpha)+ tilde(beta))=d tilde(alpha)+d tilde(beta):}\begin{equation*} \boldsymbol{d}(\tilde{\boldsymbol{\alpha}}+\tilde{\boldsymbol{\beta}})=\boldsymbol{d} \tilde{\boldsymbol{\alpha}}+\boldsymbol{d} \tilde{\boldsymbol{\beta}} \tag{33.1} \end{equation*}(33.1)d(α~+β~)=dα~+dβ~
II: For the 0 -form f f fff (i.e. a scalar function), the 1 -form d f d f df\boldsymbol{d} fdf is defined by the usual relationship
(33.2) d f , v = v [ f ] = v f (33.2) d f , v = v [ f ] = v f {:(33.2)(:df","v:)=v[f]=del_(v)f:}\begin{equation*} \langle\boldsymbol{d} f, \boldsymbol{v}\rangle=\boldsymbol{v}[f]=\partial_{\boldsymbol{v}} f \tag{33.2} \end{equation*}(33.2)df,v=v[f]=vf
where v v v\boldsymbol{v}v is a vector.
In coordinates, the 1 -form is given by
(33.3) d f = f x k d x k (33.3) d f = f x k d x k {:(33.3)df=(del f)/(delx^(k))dx^(k):}\begin{equation*} \boldsymbol{d} f=\frac{\partial f}{\partial x^{k}} \boldsymbol{d} x^{k} \tag{33.3} \end{equation*}(33.3)df=fxkdxk
III: The action of d d d\boldsymbol{d}d on a wedge product is given by
d ( α ~ β ~ ) = d α ~ β ~ + ( 1 ) p α ~ d β ~ d ( α ~ β ~ ) = d α ~ β ~ + ( 1 ) p α ~ d β ~ d( tilde(alpha)^^ tilde(beta))=d tilde(alpha)^^ tilde(beta)+(-1)^(p) tilde(alpha)^^d tilde(beta)\boldsymbol{d}(\tilde{\boldsymbol{\alpha}} \wedge \tilde{\boldsymbol{\beta}})=\boldsymbol{d} \tilde{\boldsymbol{\alpha}} \wedge \tilde{\boldsymbol{\beta}}+(-1)^{p} \tilde{\boldsymbol{\alpha}} \wedge \boldsymbol{d} \tilde{\boldsymbol{\beta}}d(α~β~)=dα~β~+(1)pα~dβ~
(33.4)
exte-
IV: The exterior derivative of an exterior derivative is zero
(33.5) d 2 = d d = 0 (33.5) d 2 = d d = 0 {:(33.5)d^(2)=dd=0:}\begin{equation*} d^{2}=d d=0 \tag{33.5} \end{equation*}(33.5)d2=dd=0
1 1 ^(1){ }^{1}1 Alternatively, we could write
d A ~ = d A ~ = d tilde(A)=\boldsymbol{d} \tilde{A}=dA~=
(33.9) A | α β σ | x μ d x μ d x α d x β d x σ (33.9) A | α β σ | x μ d x μ d x α d x β d x σ {:(33.9)(delA_(|alpha beta dots sigma|))/(delx^(mu))dx^(mu)^^dx^(alpha)^^dx^(beta)^^dots^^dx^(sigma):}\begin{equation*} \frac{\partial A_{|\alpha \beta \ldots \sigma|}}{\partial x^{\mu}} \boldsymbol{d} x^{\mu} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \wedge \ldots \wedge \boldsymbol{d} x^{\sigma} \tag{33.9} \end{equation*}(33.9)A|αβσ|xμdxμdxαdxβdxσ
where we don't require the factor 1 / p 1 / p 1//p1 / p1/p !
Example 3: Consider a 1-form A ~ = a ( t ) sin χ sin θ d ϕ A ~ = a ( t ) sin χ sin θ d ϕ tilde(A)=a(t)sin chi sin theta d phi\tilde{\boldsymbol{A}}=a(t) \sin \chi \sin \theta \boldsymbol{d} \phiA~=a(t)sinχsinθdϕ. This time we have the exterior derivative
d A ~ = a ( t ) t sin θ d t d ϕ + a ( t ) cos χ sin θ d χ d ϕ + a ( t ) sin χ cos θ d θ d ϕ d A ~ = a ( t ) t sin θ d t d ϕ + a ( t ) cos χ sin θ d χ d ϕ + a ( t ) sin χ cos θ d θ d ϕ d tilde(A)=(del a(t))/(del t)sin theta dt^^d phi+a(t)cos chi sin theta d chi^^d phi+a(t)sin chi cos theta d theta^^d phi\boldsymbol{d} \tilde{\boldsymbol{A}}=\frac{\partial a(t)}{\partial t} \sin \theta \boldsymbol{d} t \wedge \boldsymbol{d} \phi+a(t) \cos \chi \sin \theta \boldsymbol{d} \chi \wedge \boldsymbol{d} \phi+a(t) \sin \chi \cos \theta \boldsymbol{d} \theta \wedge \boldsymbol{d} \phidA~=a(t)tsinθdtdϕ+a(t)cosχsinθdχdϕ+a(t)sinχcosθdθdϕ
All of the 1 -forms chosen in this example will be met again when we examine the metric of the expanding Universe.
The rule d d = 0 d d = 0 dd=0\boldsymbol{d} \boldsymbol{d}=0dd=0 puts constraints on the sorts of objects that can be accommodated in (3+1)-dimensional spacetime, as we shall see in the next example.

Example 33.2

We work in (3+1)-dimensional spacetime. Starting with a function f f fff, we can construct a 1 -form α ~ α ~ tilde(alpha)\tilde{\boldsymbol{\alpha}}α~ using the exterior derivative
(33.16) α ~ = d f (33.16) α ~ = d f {:(33.16) tilde(alpha)=df:}\begin{equation*} \tilde{\boldsymbol{\alpha}}=\boldsymbol{d} f \tag{33.16} \end{equation*}(33.16)α~=df
The 1 -form looks like a set of evenly spaced planes. If we try to obtain a 2 -form by acting on α ~ α ~ tilde(alpha)\tilde{\boldsymbol{\alpha}}α~ with the exterior derivative we obtain
(33.17) d α ~ = d d f = 0 (33.17) d α ~ = d d f = 0 {:(33.17)d tilde(alpha)=ddf=0:}\begin{equation*} \boldsymbol{d} \tilde{\alpha}=\boldsymbol{d d} f=0 \tag{33.17} \end{equation*}(33.17)dα~=ddf=0
If we start with a 1 -form A ~ A ~ tilde(A)\tilde{\boldsymbol{A}}A~, we can construct a 2 -form by acting on the 1 -form with d, which gives
(33.18) F ~ = d A ~ (33.18) F ~ = d A ~ {:(33.18) tilde(F)=d tilde(A):}\begin{equation*} \tilde{\boldsymbol{F}}=\boldsymbol{d} \tilde{\boldsymbol{A}} \tag{33.18} \end{equation*}(33.18)F~=dA~
The 2-form looks like a set of tubes in space, with the 2 -form field circulating in each of the tubes. If we try to obtain a 3 -form from the 2 -form using the exterior derivative we obtain
(33.19) d F ~ = d d A ~ = 0 (33.19) d F ~ = d d A ~ = 0 {:(33.19)d tilde(F)=dd tilde(A)=0:}\begin{equation*} d \tilde{\boldsymbol{F}}=d d \tilde{\boldsymbol{A}}=0 \tag{33.19} \end{equation*}(33.19)dF~=ddA~=0
Now start with a 2 -form ν ~ ν ~ tilde(nu)\tilde{\boldsymbol{\nu}}ν~ and construct a 3 -form by acting on ν ~ ν ~ tilde(nu)\tilde{\boldsymbol{\nu}}ν~ with d d d\boldsymbol{d}d, so that we have
(33.20) μ ~ = d ν ~ (33.20) μ ~ = d ν ~ {:(33.20) tilde(mu)=d tilde(nu):}\begin{equation*} \tilde{\mu}=d \tilde{\nu} \tag{33.20} \end{equation*}(33.20)μ~=dν~
The 3 -form μ ~ μ ~ tilde(mu)\tilde{\mu}μ~ looks like a set of cubes with the 3 -form field circulating inside. If we try to obtain a 4 -form from the 3 -form using the exterior derivative we obtain
(33.21) d μ ~ = d d ν ~ = 0 (33.21) d μ ~ = d d ν ~ = 0 {:(33.21)d tilde(mu)=dd tilde(nu)=0:}\begin{equation*} d \tilde{\mu}=d d \tilde{\nu}=0 \tag{33.21} \end{equation*}(33.21)dμ~=ddν~=0
Continuing, we can start with a 3 -form and make a 4 -form using d d d\boldsymbol{d}d. A 5 -form, however, cannot be accommodated in our four-dimensional space.
2 2 ^(2){ }^{2}2 Marius Sophus Lie (1842-1899). His name is pronounced 'Lee' (as in Bruce Lee, to rhyme with ski). Continuous transformation groups are now called Lie groups in honour of his greatest work in mathematics, and are very widely used in quantum mechanics Élie Cartan was a doctoral student of Lie.
\square
We now turn to the second sort of derivative for coordinate-free geometry: the Lie derivative. 2 2 ^(2){ }^{2}2 This derivative is most simply applied using the notion of a commutator, so we first pause to explore this in the next section.

33.2 Commutators

Adopting the idea of a vector as an operator, we define the commutator [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] of two vector fields u = d / d λ u = d / d λ u=d//dlambda\boldsymbol{u}=\mathrm{d} / \mathrm{d} \lambdau=d/dλ and v = d / d μ v = d / d μ v=d//dmu\boldsymbol{v}=\mathrm{d} / \mathrm{d} \muv=d/dμ as
[ u , v ] = u v v u (33.22) = d d λ d d μ d d μ d d λ . [ u , v ] = u v v u (33.22) = d d λ d d μ d d μ d d λ . {:[[u","v]=uv-vu],[(33.22)=(d)/((d)lambda)*((d))/((d)mu)-(d)/((d)mu)*((d))/((d)lambda).]:}\begin{align*} {[\boldsymbol{u}, \boldsymbol{v}] } & =\boldsymbol{u} \boldsymbol{v}-\boldsymbol{v} \boldsymbol{u} \\ & =\frac{\mathrm{d}}{\mathrm{~d} \lambda} \cdot \frac{\mathrm{~d}}{\mathrm{~d} \mu}-\frac{\mathrm{d}}{\mathrm{~d} \mu} \cdot \frac{\mathrm{~d}}{\mathrm{~d} \lambda} . \tag{33.22} \end{align*}[u,v]=uvvu(33.22)=d dλ d dμd dμ d dλ.
The brackets in the notation for commutators are also known as Lie brackets. Since vector fields act on functions, we can also write
[ u , v ] [ f ] = u v [ f ] v u [ f ] (33.23) = d d λ d f d μ d d μ d f d λ . [ u , v ] [ f ] = u v [ f ] v u [ f ] (33.23) = d d λ d f d μ d d μ d f d λ . {:[[u","v][f]=uv[f]-vu[f]],[(33.23)=(d)/((d)lambda)((d)f)/((d)mu)-(d)/((d)mu)((d)f)/((d)lambda).]:}\begin{align*} {[\boldsymbol{u}, \boldsymbol{v}][f] } & =\boldsymbol{u} \boldsymbol{v}[f]-\boldsymbol{v} \boldsymbol{u}[f] \\ & =\frac{\mathrm{d}}{\mathrm{~d} \lambda} \frac{\mathrm{~d} f}{\mathrm{~d} \mu}-\frac{\mathrm{d}}{\mathrm{~d} \mu} \frac{\mathrm{~d} f}{\mathrm{~d} \lambda} . \tag{33.23} \end{align*}[u,v][f]=uv[f]vu[f](33.23)=d dλ df dμd dμ df dλ.

Example 33.3

Worked through in coordinates on a function f f fff, the commutator becomes
[ u , v ] [ f ] = u α x α ( v β f x β ) v α x α ( u β f x β ) (33.24) = ( u α v β x α v α u β x α ) f x β . [ u , v ] [ f ] = u α x α v β f x β v α x α u β f x β (33.24) = u α v β x α v α u β x α f x β . {:[[u","v][f]=u^(alpha)(del)/(delx^(alpha))(v^(beta)(del f)/(delx^(beta)))-v^(alpha)(del)/(delx^(alpha))(u^(beta)(del f)/(delx^(beta)))],[(33.24)=(u^(alpha)(delv^(beta))/(delx^(alpha))-v^(alpha)(delu^(beta))/(delx^(alpha)))(del f)/(delx^(beta)).]:}\begin{align*} {[\boldsymbol{u}, \boldsymbol{v}][f] } & =u^{\alpha} \frac{\partial}{\partial x^{\alpha}}\left(v^{\beta} \frac{\partial f}{\partial x^{\beta}}\right)-v^{\alpha} \frac{\partial}{\partial x^{\alpha}}\left(u^{\beta} \frac{\partial f}{\partial x^{\beta}}\right) \\ & =\left(u^{\alpha} \frac{\partial v^{\beta}}{\partial x^{\alpha}}-v^{\alpha} \frac{\partial u^{\beta}}{\partial x^{\alpha}}\right) \frac{\partial f}{\partial x^{\beta}} . \tag{33.24} \end{align*}[u,v][f]=uαxα(vβfxβ)vαxα(uβfxβ)(33.24)=(uαvβxαvαuβxα)fxβ.
We could write the right-hand side of this equation out as another vector field w β / x β w β / x β w^(beta)del//delx^(beta)w^{\beta} \partial / \partial x^{\beta}wβ/xβ, acting on the function f f fff. The components w β w β w^(beta)w^{\beta}wβ are given by w β = w β = w^(beta)=w^{\beta}=wβ= ( u α v β x α v α u β x α ) u α v β x α v α u β x α (u^(alpha)(delv^(beta))/(delx^(alpha))-v^(alpha)(delu^(beta))/(delx^(alpha)))\left(u^{\alpha} \frac{\partial v^{\beta}}{\partial x^{\alpha}}-v^{\alpha} \frac{\partial u^{\beta}}{\partial x^{\alpha}}\right)(uαvβxαvαuβxα), which will, in general, give is a set of non-zero numbers. We conclude that the commutator generates another vector field, whose components do not, in general, vanish.
Another conclusion that can be drawn from the previous example, through the removal of the function is that the commutator, written in terms of coordinates, is
(33.25) [ u , v ] = ( u α v β x α v α u β x α ) x β (33.25) [ u , v ] = u α v β x α v α u β x α x β {:(33.25)[u","v]=(u^(alpha)(delv^(beta))/(delx^(alpha))-v^(alpha)(delu^(beta))/(delx^(alpha)))(del)/(delx^(beta)):}\begin{equation*} [\boldsymbol{u}, \boldsymbol{v}]=\left(u^{\alpha} \frac{\partial v^{\beta}}{\partial x^{\alpha}}-v^{\alpha} \frac{\partial u^{\beta}}{\partial x^{\alpha}}\right) \frac{\partial}{\partial x^{\beta}} \tag{33.25} \end{equation*}(33.25)[u,v]=(uαvβxαvαuβxα)xβ
This expression, built from directional derivatives, gives rise to a simple pictorial view of the commutator. The first term is an instruction to evaluate the difference Δ 2 Δ 2 Delta_(2)\boldsymbol{\Delta}_{2}Δ2 in the vector field v v v\boldsymbol{v}v evaluated at the tip and base of the vector u u u\boldsymbol{u}u, as shown in Fig. 33.2(b). The second term tells us to evaluate the difference Δ 1 Δ 1 Delta_(1)\boldsymbol{\Delta}_{1}Δ1 in u u u\boldsymbol{u}u evaluated at the tip and base of the vector v v v\boldsymbol{v}v, as shown in Fig. 33.2(a). The commutator ( Δ 2 Δ 1 ) Δ 2 Δ 1 (Delta_(2)-Delta_(1))\left(\boldsymbol{\Delta}_{2}-\boldsymbol{\Delta}_{1}\right)(Δ2Δ1) therefore tells us to move along the vector v v v\boldsymbol{v}v and then along the vector u u u\boldsymbol{u}u and compare this to the final position if we move along u u u\boldsymbol{u}u and then v v v\boldsymbol{v}v. In pictures, this is a measurement of the vector that closes the figures in Fig. 33.2(c). If this were Euclidean space with constant vector fields, then the parallelogram that is made is a closed figure and the commutator is zero. However, in general, we cannot guarantee that vector fields make closed figures when transported along each other's lengths. The amount by which the figure fails to close is measured by the vector that is outputted by the commutator, as shown in Fig. 33.2(c). So, in short, the commutator measures the amount by which our attempt to make a parallelogram from vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v fails to close.
In Chapter 10, we discussed the case that the vectors in the Lie brackets are the basis vectors e μ e μ e_(mu)e_{\mu}eμ of a space. Since basis vectors should, in coordinates, be expressible simply as partial derivatives / x β / x β del//delx^(beta)\partial / \partial x^{\beta}/xβ, then we expect the part in the brackets to vanish. If it does not, it implies that
Fig. 33.2 (a) The sequence v u v u vu\boldsymbol{v} \boldsymbol{u}vu involves following v v v\boldsymbol{v}v then u u u\boldsymbol{u}u. This vector u u u\boldsymbol{u}u at the tip of v v v\boldsymbol{v}v will generally be different to the one at the base of v v v\boldsymbol{v}v. (b) The sequence u v u v uv\boldsymbol{u v}uv involves following u u u\boldsymbol{u}u and then v v v\boldsymbol{v}v. (c) The commutator [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v].
3 3 ^(3){ }^{3}3 This is just what we discussed in Chapters 3 and 10.
4 4 ^(4){ }^{4}4 Although this might all seem somewhat contrived, the process of being carried along by a vector field is an example of a diffeomorphism, which is the generalization of an active cooris the generalization of an active coor-
dinate transformation to the manifold. dinate transformation to the maniford. Diffeomorphisms allow us to describe
the smoothness of a spacetime and are important in the mathematical description of relativity. See Appendix C for more details. In terms of physics, the Lie derivative is so important to us because, in cosmology, we consider that we are all being carried along the current of the cosmological fluid of matter that fills the Universe. Fluids themselves are examined in detail in Chapter 39 .
Fig. 33.3 Moving along a congruence of curves by an amount Δ λ Δ λ Delta lambda\Delta \lambdaΔλ and comparing vectors from the field v v v\boldsymbol{v}v. The paring vectors from the field v v vvv. The
congruence is formed from the streamcongruence is formed from the stream-
lines of a vector field u u u\boldsymbol{u}u. We take Lie lines of a vector field u u u\boldsymbol{u}u. We take Lie
derivatives with respect to this second derivat
field.
5 5 ^(5){ }^{5}5 The pound sign £ £ £££ means 'L' for Lie here. It's use in denoting the currency stems from the origin of the Enrency stems from the origin of the En-
glish word pound, which comes from glish word pound, which comes from
the Latin libra pondo meaning a 'pound the Latin lib
6 6 ^(6){ }^{6}6 In Chapter 31, we used given curves to provide a tangent vector fields. Now we turn the tables and take a vector field to generate a series of curves to which the field is tangent.
it is impossible to express the basis in terms of derivatives of coordinates. Such bases are non-coordinate bases. In contrast, the set of basis vectors { e μ } e μ {e_(mu)}\left\{\boldsymbol{e}_{\mu}\right\}{eμ} is a coordinate basis if and only if [ e α , e β ] = 0 . 3 e α , e β = 0 . 3 [e_(alpha),e_(beta)]=0.^(3)\left[\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right]=0 .^{3}[eα,eβ]=0.3 In the formulation of general relativity, many expressions are simplest when referred to a coordinate basis. However, we make and interpret measurements locally in orthonormal frames, which are generally noncoordinate frames.

33.3 Lie derivatives of vectors

Previously, we saw that the exterior derivative measures the rate of change of a 1-form in the direction of the normal to the 1-form's surface. When generalized, this derivative applied only to p p ppp-forms. We now turn to an alternative, primitive, method of taking derivatives. Recall that a problem we face is that it's difficult to know how to compare tensor fields at two different points in space. Lie differentiation presents a solution to this problem by using a second vector field to provide a means of moving the original vectors around.
Imagine being carried along by the current in a river. 4 4 ^(4){ }^{4}4 In order to evaluate the change of some vector field v v v\boldsymbol{v}v, we can extract a vector at some position, allow ourselves to be carried along by the river current, and then compare the vector we originally extracted to another one at the new position at which we find ourselves. This is the essence of the Lie derivative. The current of the river can itself be represented by a second vector field u u u\boldsymbol{u}u representing the velocity field of the river fluid. The vectors from the field u u u\boldsymbol{u}u are tangent to a congruence of curves that we call the streamlines of the vector field u u u\boldsymbol{u}u. These are shown in Fig. 33.3. We need to know both v v v\boldsymbol{v}v and u u u\boldsymbol{u}u to take the Lie derivative, which will turn out simply to be given by the commutator of the two fields [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v].
Lie differentiation doesn't just apply to vectors like v v v\boldsymbol{v}v; it can be applied to any tensor Z Z Z\boldsymbol{Z}Z. We write the Lie derivative of Z as 5 Z as 5 Zas^(5)\boldsymbol{Z} \mathrm{as}^{5}Zas5
(33.26) £ u Z = ( Lie derivative of tensor Z carried along vector field u ) (33.26) £ u Z = (  Lie derivative of tensor  Z  carried along vector field  u ) {:(33.26)£_(u)Z=((" Lie derivative of tensor "Z)/(" carried along vector field "u)):}\begin{equation*} £_{\boldsymbol{u}} \boldsymbol{Z}=\binom{\text { Lie derivative of tensor } \boldsymbol{Z}}{\text { carried along vector field } \boldsymbol{u}} \tag{33.26} \end{equation*}(33.26)£uZ=( Lie derivative of tensor Z carried along vector field u)
We can think of the Lie derivative as being a derivative of Z Z Z\boldsymbol{Z}Z with respect to u u u\boldsymbol{u}u, where the field u u u\boldsymbol{u}u is used to provide a yardstick against which to measure changes in Z Z Z\boldsymbol{Z}Z. Importantly, if there is no difference in a tensor after being carried along by the current, we have £ u Z = 0 £ u Z = 0 £_(u)Z=0£_{\boldsymbol{u}} \boldsymbol{Z}=0£uZ=0 and we say that the tensor field has been Lie dragged. Geometrically, a Liedragged vector that joins two streamlines at some point will continue to join them after being carried along by the current, as shown in Fig. 33.4.
Next, we show that the Lie derivative of a vector field v v v\boldsymbol{v}v obeys £ u v = £ u v = £_(u)v=£_{\boldsymbol{u}} \boldsymbol{v}=£uv= [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v]. First, we need a suitable set of streamlines along which to carry the field. In a fluid, a streamline is a curve whose tangent at some point P P P\mathcal{P}P gives the velocity u u u\boldsymbol{u}u of the element of fluid at P P P\mathcal{P}P. So we can think of the vector field u ( x ) u ( x ) u(x)\boldsymbol{u}(x)u(x) as supplying the streamlines. 6 6 ^(6){ }^{6}6

Example 33.4

More formally, we have a congruence when there is a unique curve c ( λ ) c ( λ ) c(lambda)c(\lambda)c(λ) through each point P P P\mathcal{P}P of space such that: (i) c ( λ = 0 ) = P c ( λ = 0 ) = P c(lambda=0)=Pc(\lambda=0)=\mathcal{P}c(λ=0)=P, and (ii) the tangent vector at a point c ( λ ) c ( λ ) c(lambda)c(\lambda)c(λ) is u u u quad\boldsymbol{u} \quadu This curve therefore has a tangent vector that is always given by the c ( λ ) col c ( λ ) c ( λ ) col c ( λ ) c(lambda)col_(c)(lambda)c(\lambda) \operatorname{col}_{c}(\lambda)c(λ)colc(λ). The curve is called an integral curve of the vector field u u uuu. by the obvious that such a curve will exist for some given field u u u\boldsymbol{u}u, so we pause to consider it. The components of the vector field u u u\boldsymbol{u}u at P P P\mathcal{P}P are u μ ( P ) u μ ( P ) u^(mu)(P)u^{\mu}(\mathcal{P})uμ(P) and change as a function it. The components of the vector field u u u\boldsymbol{u}u at P P P\mathcal{P}P are u μ ( P ) u μ ( P ) u^(mu)(P)u^{\mu}(\mathcal{P})uμ(P) and change as a function of this position. If we work in a coordinate system the curve has coordinates x μ ( λ ) x μ ( λ ) x^(mu)(lambda)x^{\mu}(\lambda)xμ(λ). Saying that the components u μ ( x 1 ( λ ) , x 2 ( λ ) , x n ( λ ) ) u μ x 1 ( λ ) , x 2 ( λ ) , x n ( λ ) u^(mu)(x^(1)(lambda),x^(2)(lambda),dotsx^(n)(lambda))u^{\mu}\left(x^{1}(\lambda), x^{2}(\lambda), \ldots x^{n}(\lambda)\right)uμ(x1(λ),x2(λ),xn(λ)) are tangent to the curve with parameter λ λ lambda\lambdaλ amounts to the set of differential equations
(33.27) d x μ ( λ ) d λ = u μ ( x 1 ( λ ) , x 2 ( λ ) , , x n ( λ ) ) . (33.27) d x μ ( λ ) d λ = u μ x 1 ( λ ) , x 2 ( λ ) , , x n ( λ ) . {:(33.27)(dx^(mu)(lambda))/(dlambda)=u^(mu)(x^(1)(lambda),x^(2)(lambda),dots,x^(n)(lambda)).:}\begin{equation*} \frac{\mathrm{d} x^{\mu}(\lambda)}{\mathrm{d} \lambda}=u^{\mu}\left(x^{1}(\lambda), x^{2}(\lambda), \ldots, x^{n}(\lambda)\right) . \tag{33.27} \end{equation*}(33.27)dxμ(λ)dλ=uμ(x1(λ),x2(λ),,xn(λ)).
This set of first-order differential equations is guaranteed 7 7 ^(7){ }^{7}7 to have unique solutions in the region near P P P\mathcal{P}P. This result guarantees the existence of the required congruence of integral curves.
With all of the ingredients in place, we're now ready to compute the derivative. We take the vector field at P P P\mathcal{P}P, called v ( P ) v ( P ) v(P)\boldsymbol{v}(\mathcal{P})v(P), we carry this vector a distance Δ λ Δ λ Delta lambda\Delta \lambdaΔλ along an integral curve of the vector field u u u\boldsymbol{u}u to a point Q Q Q\mathcal{Q}Q. This turns v ( P ) v ( P ) v(P)\boldsymbol{v}(\mathcal{P})v(P) into the vector v ( Q ) v ( Q ) v^(')(Q)\boldsymbol{v}^{\prime}(\mathcal{Q})v(Q). We compare this transported vector to the vector v ( Q ) v ( Q ) v(Q)\boldsymbol{v}(\mathcal{Q})v(Q) found at Q Q Q\mathcal{Q}Q. The definition of the Lie derivative is then
(33.28) £ u v ( P ) = lim Δ λ 0 1 Δ λ { v ( Q ) v ( Q ) } (33.28) £ u v ( P ) = lim Δ λ 0 1 Δ λ v ( Q ) v ( Q ) {:(33.28)£_(u)v(P)=lim_(Delta lambda rarr0)(1)/(Delta lambda){v(Q)-v^(')(Q)}:}\begin{equation*} £_{\boldsymbol{u}} \boldsymbol{v}(\mathcal{P})=\lim _{\Delta \lambda \rightarrow 0} \frac{1}{\Delta \lambda}\left\{\boldsymbol{v}(\mathcal{Q})-\boldsymbol{v}^{\prime}(\mathcal{Q})\right\} \tag{33.28} \end{equation*}(33.28)£uv(P)=limΔλ01Δλ{v(Q)v(Q)}
In the next example, we carry out this computation in a coordinate system.

Example 33.5

In terms of a coordinate system, we have a curve parametrized by λ λ lambda\lambdaλ with a tangent vector with components u μ = d x μ d λ u μ = d x μ d λ u^(mu)=(dx^(mu))/(d lambda)u^{\mu}=\frac{d x^{\mu}}{d \lambda}uμ=dxμdλ. The point P P P\mathcal{P}P has a coordinate x μ x μ x^(mu)x^{\mu}xμ. The coordinates of the nearby point Q Q Q\mathcal{Q}Q are x μ x μ x^(mu^('))x^{\mu^{\prime}}xμ and these can be found using a coordinate transformation that translates us from P P P\mathcal{P}P to Q Q Q\mathcal{Q}Q, along the vector u u u\boldsymbol{u}u
(33.29) x μ = x μ + d x μ = x μ + u μ d λ . (33.29) x μ = x μ + d x μ = x μ + u μ d λ . {:(33.29)x^(mu^('))=x^(mu)+dx^(mu)=x^(mu)+u^(mu)dlambda.:}\begin{equation*} x^{\mu^{\prime}}=x^{\mu}+\mathrm{d} x^{\mu}=x^{\mu}+u^{\mu} \mathrm{d} \lambda . \tag{33.29} \end{equation*}(33.29)xμ=xμ+dxμ=xμ+uμdλ.
To find the Lie derivative, we need to compare the field at a point Q Q Q\mathcal{Q}Q (with components v β v β v^(beta)v^{\beta}vβ ) to the field carried from P P P\mathcal{P}P to Q Q Q\mathcal{Q}Q (with components v β v β v^(beta^('))v^{\beta^{\prime}}vβ )
(33.30) ( £ u v ( P ) ) β = v β ( Q ) v β ( Q ) d λ (33.30) £ u v ( P ) β = v β ( Q ) v β ( Q ) d λ {:(33.30)(£_(u)v(P))^(beta)=(v^(beta)(Q)-v^(beta^('))(Q))/(dlambda):}\begin{equation*} \left(£_{\boldsymbol{u}} \boldsymbol{v}(\mathcal{P})\right)^{\beta}=\frac{v^{\beta}(\mathcal{Q})-v^{\beta^{\prime}}(\mathcal{Q})}{\mathrm{d} \lambda} \tag{33.30} \end{equation*}(33.30)(£uv(P))β=vβ(Q)vβ(Q)dλ
Step I: We first work out the components of the transported vector. We do this by transforming the vector into the coordinate frame at Q Q Q\mathcal{Q}Q using the tensor transformation law 8 8 ^(8){ }^{8}8 and then substituting eqn 33.29, thus
v β ( Q ) = x β x α v α ( P ) = ( δ β α + u β x α d λ ) v α ( P ) (33.31) = v β ( P ) + u β x α v α ( P ) d λ . v β ( Q ) = x β x α v α ( P ) = δ β α + u β x α d λ v α ( P ) (33.31) = v β ( P ) + u β x α v α ( P ) d λ . {:[v^(beta^('))(Q)=(delx^(beta^(')))/(delx^(alpha))v^(alpha)(P)],[=(delta^(beta)_(alpha)+(delu^(beta))/(delx^(alpha))*dlambda)v^(alpha)(P)],[(33.31)=v^(beta)(P)+(delu^(beta))/(delx^(alpha))v^(alpha)(P)dlambda.]:}\begin{align*} v^{\beta^{\prime}}(\mathcal{Q}) & =\frac{\partial x^{\beta^{\prime}}}{\partial x^{\alpha}} v^{\alpha}(\mathcal{P}) \\ & =\left(\delta^{\beta}{ }_{\alpha}+\frac{\partial u^{\beta}}{\partial x^{\alpha}} \cdot \mathrm{d} \lambda\right) v^{\alpha}(\mathcal{P}) \\ & =v^{\beta}(\mathcal{P})+\frac{\partial u^{\beta}}{\partial x^{\alpha}} v^{\alpha}(\mathcal{P}) \mathrm{d} \lambda . \tag{33.31} \end{align*}vβ(Q)=xβxαvα(P)=(δβα+uβxαdλ)vα(P)(33.31)=vβ(P)+uβxαvα(P)dλ.
Fig. 33.4 A Lie dragged vector continues to link two streamlines after being carried along the congruence.
7 7 ^(7){ }^{7}7 This is a result which is proved in many of standard references. An example is Choquet-Bruhat, DeWittMorette and Dillard-Bleick Analysis, Manifolds and Physics.


Fig. 33.5 (a) The effect of transporting the vector v v v\boldsymbol{v}v from P P P\mathcal{P}P to Q Q Q\mathcal{Q}Q is measured in terms of the change in u u u\boldsymbol{u}u. (b) The vector v v v\boldsymbol{v}v evaluated at P P P\mathcal{P}P and Q Q Q\mathcal{Q}Q. (c) The difference between (a) and (b) (c) The difference between (a) and (b) gives the Lie derivative: a vector that
closes the circuit. closes the circuit.
8 8 ^(8){ }^{8}8 Remember that components must transform as A μ = x μ x ν A ν A μ = x μ x ν A ν A^(mu^('))=(delx^(mu^(')))/(delx^(nu))A^(nu)A^{\mu^{\prime}}=\frac{\partial x^{\mu^{\prime}}}{\partial x^{\nu}} A^{\nu}Aμ=xμxνAν.
9 9 ^(9){ }^{9}9 Note that if we have a connection Γ β α γ Γ β α γ Gamma^(beta)_(alpha gamma)\Gamma^{\beta}{ }_{\alpha \gamma}Γβαγ, this expression can be rewritten in terms of covariant derivatives in an equivalent form
( £ u v ( P ) ) β = u α ( v β x α + Γ β α γ v γ ) v α ( u β x α + Γ β α γ u γ ) (33.33) = u α v β ; α v α u β ; α . £ u v ( P ) β = u α v β x α + Γ β α γ v γ v α u β x α + Γ β α γ u γ (33.33) = u α v β ; α v α u β ; α . {:[(£_(u)v(P))^(beta)=u^(alpha)((delv^(beta))/(delx^(alpha))+Gamma^(beta)_(alpha gamma)v^(gamma))],[-v^(alpha)((delu^(beta))/(delx^(alpha))+Gamma^(beta)_(alpha gamma)u^(gamma))],[(33.33)=u^(alpha)v^(beta)_(;alpha)-v^(alpha)u^(beta)_(;alpha).]:}\begin{align*} \left(£_{\boldsymbol{u}} \boldsymbol{v}(\mathcal{P})\right)^{\beta}= & u^{\alpha}\left(\frac{\partial v^{\beta}}{\partial x^{\alpha}}+\Gamma^{\beta}{ }_{\alpha \gamma} v^{\gamma}\right) \\ & -v^{\alpha}\left(\frac{\partial u^{\beta}}{\partial x^{\alpha}}+\Gamma^{\beta}{ }_{\alpha \gamma} u^{\gamma}\right) \\ = & u^{\alpha} v^{\beta}{ }_{; \alpha}-v^{\alpha} u^{\beta}{ }_{; \alpha} . \tag{33.33} \end{align*}(£uv(P))β=uα(vβxα+Γβαγvγ)vα(uβxα+Γβαγuγ)(33.33)=uαvβ;αvαuβ;α.
This is strictly optional: the connection is not needed to define the Lie derivative.
We see that the effect of carrying the vector v v v\boldsymbol{v}v is measured by u β x α v α ( P ) u β x α v α ( P ) (delu^(beta))/(delx^(alpha))*v^(alpha)(P)\frac{\partial u^{\beta}}{\partial x^{\alpha}} \cdot v^{\alpha}(\mathcal{P})uβxαvα(P), which is the difference in the velocity field u u u\boldsymbol{u}u measured at the base and tip of the v v v\boldsymbol{v}v, as shown in Fig. 33.5(a)
Step II: Second, we compute the vector v v v\boldsymbol{v}v at Q Q Q\mathcal{Q}Q with components v β ( Q ) v β ( Q ) v^(beta)(Q)v^{\beta}(\mathcal{Q})vβ(Q). This is done by seeing how the field v v v\boldsymbol{v}v itself changes between P P P\mathcal{P}P and Q Q Q\mathcal{Q}Q as we move along u u u\boldsymbol{u}u
(33.32) v β ( Q ) = v β ( P ) + v β x α u α ( P ) d λ (33.32) v β ( Q ) = v β ( P ) + v β x α u α ( P ) d λ {:(33.32)v^(beta)(Q)=v^(beta)(P)+(delv^(beta))/(delx^(alpha))u^(alpha)(P)dlambda:}\begin{equation*} v^{\beta}(\mathcal{Q})=v^{\beta}(\mathcal{P})+\frac{\partial v^{\beta}}{\partial x^{\alpha}} u^{\alpha}(\mathcal{P}) \mathrm{d} \lambda \tag{33.32} \end{equation*}(33.32)vβ(Q)=vβ(P)+vβxαuα(P)dλ
This is shown geometrically in Fig. 33.5(b).
Step III: The previous two expressions involve the field evaluated at the same point, so can be combined. The Lie derivative is, therefore 9 9 ^(9){ }^{9}9
( £ u v ( P ) ) β = v β ( Q ) v β ( Q ) d λ . (33.34) = u α ( P ) v β x α v α ( P ) u β x α £ u v ( P ) β = v β ( Q ) v β ( Q ) d λ . (33.34) = u α ( P ) v β x α v α ( P ) u β x α {:[(£_(u)v(P))^(beta)=(v^(beta)(Q)-v^(beta^('))(Q))/(dlambda).],[(33.34)=u^(alpha)(P)(delv^(beta))/(delx^(alpha))-v^(alpha)(P)(delu^(beta))/(delx^(alpha))]:}\begin{align*} \left(£_{\boldsymbol{u}} \boldsymbol{v}(\mathcal{P})\right)^{\beta} & =\frac{v^{\beta}(\mathcal{Q})-v^{\beta^{\prime}}(\mathcal{Q})}{\mathrm{d} \lambda} . \\ & =u^{\alpha}(\mathcal{P}) \frac{\partial v^{\beta}}{\partial x^{\alpha}}-v^{\alpha}(\mathcal{P}) \frac{\partial u^{\beta}}{\partial x^{\alpha}} \tag{33.34} \end{align*}(£uv(P))β=vβ(Q)vβ(Q)dλ.(33.34)=uα(P)vβxαvα(P)uβxα
Comparing with eqn 33.25 , we see that this is given simply by the Lie bracket of the fields u u u\boldsymbol{u}u and v v v\boldsymbol{v}v.
The result from the last example implies that, for a vector field, we have
(33.35) £ u v = [ u , v ] . (33.35) £ u v = [ u , v ] . {:(33.35)£_(u)v=[u","v].:}\begin{equation*} £_{\boldsymbol{u}} \boldsymbol{v}=[\boldsymbol{u}, \boldsymbol{v}] . \tag{33.35} \end{equation*}(33.35)£uv=[u,v].
We use this to formalize our set of rules for finding the Lie derivative of a field.
The Lie derivative of a scalar function with respect to the field u u u\boldsymbol{u}u is defined to be the directional derivative along u u u\boldsymbol{u}u :
(33.36) £ u f = u [ f ] = u f . (33.36) £ u f = u [ f ] = u f . {:(33.36)£_(u)f=u[f]=del_(u)f.:}\begin{equation*} £_{\boldsymbol{u}} f=\boldsymbol{u}[f]=\partial_{\boldsymbol{u}} f . \tag{33.36} \end{equation*}(33.36)£uf=u[f]=uf.
The Lie derivative of a vector field v = d / d μ v = d / d μ v=d//dmu\boldsymbol{v}=\mathrm{d} / \mathrm{d} \muv=d/dμ with respect to u = d / d λ u = d / d λ u=d//dlambda\boldsymbol{u}=\mathrm{d} / \mathrm{d} \lambdau=d/dλ is
(33.37) £ u v = d d λ v d d μ u = [ u , v ] = ( u α v β , α v α u β , α ) x β . (33.37) £ u v = d d λ v d d μ u = [ u , v ] = u α v β , α v α u β , α x β . {:(33.37)£_(u)v=(d)/((d)lambda)v-(d)/((d)mu)u=[u","v]=(u^(alpha)v^(beta)_(,alpha)-v^(alpha)u^(beta)_(,alpha))(del)/(delx^(beta)).:}\begin{equation*} £_{\boldsymbol{u}} \boldsymbol{v}=\frac{\mathrm{d}}{\mathrm{~d} \lambda} \boldsymbol{v}-\frac{\mathrm{d}}{\mathrm{~d} \mu} \boldsymbol{u}=[\boldsymbol{u}, \boldsymbol{v}]=\left(u^{\alpha} v^{\beta}{ }_{, \alpha}-v^{\alpha} u^{\beta}{ }_{, \alpha}\right) \frac{\partial}{\partial x^{\beta}} . \tag{33.37} \end{equation*}(33.37)£uv=d dλvd dμu=[u,v]=(uαvβ,αvαuβ,α)xβ.
Notice from the geometrical interpretation of the Lie derivative that £ u v £ u v £_(u)v£_{\boldsymbol{u}} \boldsymbol{v}£uv evaluated at point P P P\mathcal{P}P does not just depend on the value of v v v\boldsymbol{v}v at P P P\mathcal{P}P. It also depends on the value of v v v\boldsymbol{v}v at surrounding points. This is unlike the ordinary partial derivative, which only depends on the point P P P\mathcal{P}P in question. As a result, the Lie derivative is not powerful enough to become the derivative for use describing the curvature in manifolds. This is the reason why, ultimately, we must introduce a connection and a covariant derivative.
Example 33.6
Choose a coordinate system where u u u\boldsymbol{u}u is a coordinate basis vector e 1 = / x 1 e 1 = / x 1 e_(1)=del//delx^(1)\boldsymbol{e}_{1}=\partial / \partial x^{1}e1=/x1, then we have
(33.38) ( £ e 1 W ) μ = δ ν 1 x ν W μ W ν x ν δ μ ν = W μ x 1 . (33.38) £ e 1 W μ = δ ν 1 x ν W μ W ν x ν δ μ ν = W μ x 1 . {:(33.38)(£_(e_(1))W)^(mu)=delta^(nu)_(1)(del)/(delx^(nu))W^(mu)-W^(nu)(del)/(delx^(nu))*delta^(mu)_(nu)=(delW^(mu))/(delx^(1)).:}\begin{equation*} \left(£_{\boldsymbol{e}_{1}} \boldsymbol{W}\right)^{\mu}=\delta^{\nu}{ }_{1} \frac{\partial}{\partial x^{\nu}} W^{\mu}-W^{\nu} \frac{\partial}{\partial x^{\nu}} \cdot \delta^{\mu}{ }_{\nu}=\frac{\partial W^{\mu}}{\partial x^{1}} . \tag{33.38} \end{equation*}(33.38)(£e1W)μ=δν1xνWμWνxνδμν=Wμx1.
This suggests a sense in which we can interpret the Lie derivative as a coordinate-free version of the partial derivative.

33.4 Lie derivatives of tensors

We advertised the Lie derivative as working for all tensors. Another object we need to know how to differentiate in this way is therefore a 1 -form. To find this, consider the contraction σ ~ , w σ ~ , w (: tilde(sigma),w:)\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{w}\rangleσ~,w and take the Lie derivative with respect to the vector field v v v\boldsymbol{v}v. To do this, we simply use the Leibniz rule 10 £ v σ ~ , w = £ v σ ~ , w + σ ~ , £ v w 10 £ v σ ~ , w = £ v σ ~ , w + σ ~ , £ v w ^(10)£_(v)(: tilde(sigma),w:)=(:£_(v)( tilde(sigma)),w:)+(:( tilde(sigma)),£_(v)w:){ }^{10} £_{\boldsymbol{v}}\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{w}\rangle=\left\langle £_{\boldsymbol{v}} \tilde{\boldsymbol{\sigma}}, \boldsymbol{w}\right\rangle+\left\langle\tilde{\boldsymbol{\sigma}}, £_{\boldsymbol{v}} \boldsymbol{w}\right\rangle10£vσ~,w=£vσ~,w+σ~,£vw. Since, for fields, σ ~ , w = f σ ~ , w = f (: tilde(sigma),w:)=f\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{w}\rangle=fσ~,w=f, where f f fff is some scalar field, we have £ v σ ~ , w = v f £ v σ ~ , w = v f £_(v)(: tilde(sigma),w:)=del_(v)f£_{\boldsymbol{v}}\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{w}\rangle=\partial_{\boldsymbol{v}} f£vσ~,w=vf and a result
(33.41) £ v σ ~ , w = v f σ ~ , £ v w (33.41) £ v σ ~ , w = v f σ ~ , £ v w {:(33.41)(:£_(v)( tilde(sigma)),w:)=del_(v)f-(:( tilde(sigma)),£_(v)w:):}\begin{equation*} \left\langle £_{\boldsymbol{v}} \tilde{\boldsymbol{\sigma}}, \boldsymbol{w}\right\rangle=\partial_{\boldsymbol{v}} f-\left\langle\tilde{\boldsymbol{\sigma}}, £_{\boldsymbol{v}} \boldsymbol{w}\right\rangle \tag{33.41} \end{equation*}(33.41)£vσ~,w=vfσ~,£vw
Example 33.7
From the previous expression we can, with some relabelling of indices, deduce the components of the Lie derivative of a 1 -form
( £ v σ ~ ) μ w μ = v ν x ν ( σ μ w μ ) σ μ ( v ν x ν w μ w ν x ν v μ ) (33.42) = ( v ν x ν σ μ + σ ν x μ v ν ) w μ . £ v σ ~ μ w μ = v ν x ν σ μ w μ σ μ v ν x ν w μ w ν x ν v μ (33.42) = v ν x ν σ μ + σ ν x μ v ν w μ . {:[(£_(v)( tilde(sigma)))_(mu)w^(mu)=v^(nu)(del)/(delx^(nu))(sigma_(mu)w^(mu))-sigma_(mu)(v^(nu)(del)/(delx^(nu))w^(mu)-w^(nu)(del)/(delx^(nu))v^(mu))],[(33.42)=(v^(nu)(del)/(delx^(nu))sigma_(mu)+sigma_(nu)(del)/(delx^(mu))v^(nu))w^(mu).]:}\begin{align*} \left(£_{v} \tilde{\boldsymbol{\sigma}}\right)_{\mu} w^{\mu} & =v^{\nu} \frac{\partial}{\partial x^{\nu}}\left(\sigma_{\mu} w^{\mu}\right)-\sigma_{\mu}\left(v^{\nu} \frac{\partial}{\partial x^{\nu}} w^{\mu}-w^{\nu} \frac{\partial}{\partial x^{\nu}} v^{\mu}\right) \\ & =\left(v^{\nu} \frac{\partial}{\partial x^{\nu}} \sigma_{\mu}+\sigma_{\nu} \frac{\partial}{\partial x^{\mu}} v^{\nu}\right) w^{\mu} . \tag{33.42} \end{align*}(£vσ~)μwμ=vνxν(σμwμ)σμ(vνxνwμwνxνvμ)(33.42)=(vνxνσμ+σνxμvν)wμ.
Since the w μ w μ w^(mu)w^{\mu}wμ are arbitrary, we conclude
(33.43) ( £ v σ ~ ) μ = v ν x ν σ μ + σ ν x μ v ν . (33.43) £ v σ ~ μ = v ν x ν σ μ + σ ν x μ v ν . {:(33.43)(£_(v)( tilde(sigma)))_(mu)=v^(nu)(del)/(delx^(nu))sigma_(mu)+sigma_(nu)(del)/(delx^(mu))v^(nu).:}\begin{equation*} \left(£_{\boldsymbol{v}} \tilde{\boldsymbol{\sigma}}\right)_{\mu}=v^{\nu} \frac{\partial}{\partial x^{\nu}} \sigma_{\mu}+\sigma_{\nu} \frac{\partial}{\partial x^{\mu}} v^{\nu} . \tag{33.43} \end{equation*}(33.43)(£vσ~)μ=vνxνσμ+σνxμvν.
If we want to work out Lie derivatives of larger tensor objects, we again use the Leibniz rule and find a more general expression
£ v T ( σ ~ , ; U ) = ( £ v T ) ( σ ~ , ; U ) + T ( £ v σ ~ , ; U ) + (33.44) + T ( σ ~ , ; £ v U ) + £ v T ( σ ~ , ; U ) = £ v T ( σ ~ , ; U ) + T £ v σ ~ , ; U + (33.44) + T σ ~ , ; £ v U + {:[£_(v)T( tilde(sigma)","dots;U dots)=(£_(v)T)( tilde(sigma)","dots;U dots)+T(£_(v)( tilde(sigma)),dots;U dots)+dots],[(33.44)+T(( tilde(sigma)),dots;£_(v)U dots)+dots]:}\begin{align*} £_{\boldsymbol{v}} \boldsymbol{T}(\tilde{\boldsymbol{\sigma}}, \ldots ; \boldsymbol{U} \ldots) & =\left(£_{\boldsymbol{v}} \boldsymbol{T}\right)(\tilde{\boldsymbol{\sigma}}, \ldots ; \boldsymbol{U} \ldots)+\boldsymbol{T}\left(£_{\boldsymbol{v}} \tilde{\boldsymbol{\sigma}}, \ldots ; \boldsymbol{U} \ldots\right)+\ldots \\ & +\boldsymbol{T}\left(\tilde{\boldsymbol{\sigma}}, \ldots ; £_{\boldsymbol{v}} \boldsymbol{U} \ldots\right)+\ldots \tag{33.44} \end{align*}£vT(σ~,;U)=(£vT)(σ~,;U)+T(£vσ~,;U)+(33.44)+T(σ~,;£vU)+
Perhaps most useful here is the general expression in component form
( £ v T ) ν ξ ω α β κ = v μ ( T ν ξ ω α β κ x μ ) T ν ξ ω μ β κ v α x μ T ν ξ ω α μ κ v β x μ T ν ξ ω α β μ v κ x μ (33.45) + T μ ξ ω α β κ v μ x ν + T ν μ ω α β κ v μ x ξ + T ν ξ μ α β κ v μ x ω £ v T ν ξ ω α β κ = v μ T ν ξ ω α β κ x μ T ν ξ ω μ β κ v α x μ T ν ξ ω α μ κ v β x μ T ν ξ ω α β μ v κ x μ (33.45) + T μ ξ ω α β κ v μ x ν + T ν μ ω α β κ v μ x ξ + T ν ξ μ α β κ v μ x ω {:[(£_(v)T)_(nu xi dots omega)^(alpha beta dots kappa)=v^(mu)((delT_(nu xi dots omega)^(alpha beta dots kappa))/(delx^(mu)))-T_(nu xi dots omega)^(mu beta dots kappa)(delv^(alpha))/(delx^(mu))-T_(nu xi dots omega)^(alpha mu dots kappa)(delv^(beta))/(delx^(mu)dots)-T_(nu xi dots omega)^(alpha beta dots mu)(delv^(kappa))/(delx^(mu))],[(33.45)+T_(mu xi dots omega)^(alpha beta dots kappa)(delv^(mu))/(delx^(nu))+T_(nu mu dots omega)^(alpha beta dots kappa)(delv^(mu))/(delx^(xi))dots+T_(nu xi dots mu)^(alpha beta dots kappa)(delv^(mu))/(delx^(omega))]:}\begin{align*} \left(£_{\boldsymbol{v}} \boldsymbol{T}\right)_{\nu \xi \ldots \omega}^{\alpha \beta \ldots \kappa}= & v^{\mu}\left(\frac{\partial T_{\nu \xi \ldots \omega}^{\alpha \beta \ldots \kappa}}{\partial x^{\mu}}\right)-T_{\nu \xi \ldots \omega}^{\mu \beta \ldots \kappa} \frac{\partial v^{\alpha}}{\partial x^{\mu}}-T_{\nu \xi \ldots \omega}^{\alpha \mu \ldots \kappa} \frac{\partial v^{\beta}}{\partial x^{\mu} \ldots}-T_{\nu \xi \ldots \omega}^{\alpha \beta \ldots \mu} \frac{\partial v^{\kappa}}{\partial x^{\mu}} \\ & +T_{\mu \xi \ldots \omega}^{\alpha \beta \ldots \kappa} \frac{\partial v^{\mu}}{\partial x^{\nu}}+T_{\nu \mu \ldots \omega}^{\alpha \beta \ldots \kappa} \frac{\partial v^{\mu}}{\partial x^{\xi}} \ldots+T_{\nu \xi \ldots \mu}^{\alpha \beta \ldots \kappa} \frac{\partial v^{\mu}}{\partial x^{\omega}} \tag{33.45} \end{align*}(£vT)νξωαβκ=vμ(Tνξωαβκxμ)TνξωμβκvαxμTνξωαμκvβxμTνξωαβμvκxμ(33.45)+Tμξωαβκvμxν+Tνμωαβκvμxξ+Tνξμαβκvμxω
Notice the pleasing way in which the indices behave, generalizing the rules for the up and down components.

Example 33.8

A very useful result, which follows from eqn 33.45, is that the components of the Lie derivative of the metric 2 -form are given by
(33.46) ( £ v g ) α β = v γ x γ g α β + g γ β x α v γ + g α γ x β v γ . (33.46) £ v g α β = v γ x γ g α β + g γ β x α v γ + g α γ x β v γ . {:(33.46)(£_(v)g)_(alpha beta)=v^(gamma)(del)/(delx^(gamma))g_(alpha beta)+g_(gamma beta)(del)/(delx^(alpha))v^(gamma)+g_(alpha gamma)(del)/(delx^(beta))v^(gamma).:}\begin{equation*} \left(£_{\boldsymbol{v}} \boldsymbol{g}\right)_{\alpha \beta}=v^{\gamma} \frac{\partial}{\partial x^{\gamma}} g_{\alpha \beta}+g_{\gamma \beta} \frac{\partial}{\partial x^{\alpha}} v^{\gamma}+g_{\alpha \gamma} \frac{\partial}{\partial x^{\beta}} v^{\gamma} . \tag{33.46} \end{equation*}(33.46)(£vg)αβ=vγxγgαβ+gγβxαvγ+gαγxβvγ.
del\partial

\curvearrowright Section 33.4 gives the technical recipe for applying the Lie derivative to tensors. It can be skipped if you're happy to take the useful expression in eqn 33.46 on trust.

10 10 ^(10){ }^{10}10 Using the linearity of the derivative operation, we have for arbitrary tensors, that
£ v ( S T ) = £ v S T + S £ v T £ v ( S T ) = £ v S T + S £ v T £_(v)(S ox T)=£_(v)S ox T+S ox£_(v)T£_{v}(S \otimes T)=£_{v} S \otimes T+S \otimes £_{v} T£v(ST)=£vST+S£vT. (33.39) If we contract this, we obtain a Leibniz rule
£ v S , T = £ v S , T + S , £ v T £ v S , T = £ v S , T + S , £ v T £_(v)(:S,T:)=(:£_(v)S,T:)+(:S,£_(v)T:)£_{\boldsymbol{v}}\langle\boldsymbol{S}, \boldsymbol{T}\rangle=\left\langle £_{v} \boldsymbol{S}, \boldsymbol{T}\right\rangle+\left\langle\boldsymbol{S}, £_{\boldsymbol{v}} \boldsymbol{T}\right\rangle£vS,T=£vS,T+S,£vT.
(33.40)
example of the use of Killing vectors. Noether's theorem is viscussed in Chapter 40 .
11 11 ^(11){ }^{11}11 Wilhelm Killing (1847-1923). Killing invented Lie algebras, independently of Sophus Lie, who was dismissive of Killing's work, claiming that he (Lie) had proven all that was valid in the had proven all that was valid in the
subject, while all that was invalid was subject, while all that was invalid was
added by Killing. Killing was very added by Killing. Killing was very
modest about his own work, which was modest about his own work, which was
arguably intended to be more general than Lie's. In it, Killing made a number of unproven conjectures, which only later turned out to be true.
12 12 ^(12){ }^{12}12 Note that Killing's equation can also be written in the memorable shorthand form ξ ( α ; β ) = 0 ξ ( α ; β ) = 0 xi_((alpha;beta))=0\xi_{(\alpha ; \beta)}=0ξ(α;β)=0.
After all of this setting up, we should mention some uses of the Lie derivative. The Lie derivative is the natural derivative for expressing the invariance of a tensor under a change in position along a curve. The Lie derivative arises most often in relativity in cases where we imagine ourselves carried along a particular world line while making measurements. We often use comoving coordinates in which we float along with an element of fluid, as in cosmology, for example. Another use arises when examining geodesic deviation. The vector linking the two geodesics has zero Lie derivative along the velocity field of the geodesics.

33.5 Killing vectors

A particularly important class of vectors are those that represent conserved quantities. In mechanics and field theory, Noether's theorem tells us that wherever there's a symmetry then we have a conserved quantity. There is also a geometric method of extracting the conserved quantities from the metric that relies on the use of the Lie derivative and which is particularly useful in relativity.
Geometrically, conserved quantities can be found by identifying Killing vectors. 11 11 ^(11){ }^{11}11 We say that a geometry has an isometry if we can identify a vector field ξ ξ xi\boldsymbol{\xi}ξ with the property that if a set of points is displaced along the streamlines of ξ ξ xi\boldsymbol{\xi}ξ, then all distance relationships are unchanged. This vector field ξ ξ xi\boldsymbol{\xi}ξ is then a Killing vector field for the geometry. Distance relationships are encoded by the metric tensor, and so we want the metric tensor to remain unchanged as we carry it along the streamlines of ξ ξ xi\boldsymbol{\xi}ξ. This implies we have
(33.47) £ ξ g = 0 , (33.47) £ ξ g = 0 , {:(33.47)£_(xi)g=0",":}\begin{equation*} £_{\xi} \boldsymbol{g}=0, \tag{33.47} \end{equation*}(33.47)£ξg=0,
and so, if the metric tensor is Lie dragged along the congruence formed by the field ξ ξ xi\boldsymbol{\xi}ξ, then ξ ξ xi\boldsymbol{\xi}ξ is a Killing field.
When we have access to a connection, the condition £ ξ g = 0 £ ξ g = 0 £_(xi)g=0£_{\boldsymbol{\xi}} \boldsymbol{g}=0£ξg=0 implies that ξ ξ xi\boldsymbol{\xi}ξ satisfies Killing's equation 12 12 ^(12){ }^{12}12
(33.48) ξ α ; β + ξ β ; α = 0 . (33.48) ξ α ; β + ξ β ; α = 0 . {:(33.48)xi_(alpha;beta)+xi_(beta;alpha)=0.:}\begin{equation*} \xi_{\alpha ; \beta}+\xi_{\beta ; \alpha}=0 . \tag{33.48} \end{equation*}(33.48)ξα;β+ξβ;α=0.

Example 33.9

We pause to prove ξ ( α ; β ) = 0 ξ ( α ; β ) = 0 xi_((alpha;beta))=0\xi_{(\alpha ; \beta)}=0ξ(α;β)=0 using the result in example 33.8. Write the defining equation in coordinates
(33.49) ( £ ξ g ) α β = ξ γ x γ g α β + g γ β x α ξ γ + g α γ x β ξ γ = 0 . (33.49) £ ξ g α β = ξ γ x γ g α β + g γ β x α ξ γ + g α γ x β ξ γ = 0 . {:(33.49)(£_(xi)g)_(alpha beta)=xi^(gamma)(del)/(delx^(gamma))g_(alpha beta)+g_(gamma beta)(del)/(delx^(alpha))xi^(gamma)+g_(alpha gamma)(del)/(delx^(beta))xi^(gamma)=0.:}\begin{equation*} \left(£_{\boldsymbol{\xi}} \boldsymbol{g}\right)_{\alpha \beta}=\xi^{\gamma} \frac{\partial}{\partial x^{\gamma}} g_{\alpha \beta}+g_{\gamma \beta} \frac{\partial}{\partial x^{\alpha}} \xi^{\gamma}+g_{\alpha \gamma} \frac{\partial}{\partial x^{\beta}} \xi^{\gamma}=0 . \tag{33.49} \end{equation*}(33.49)(£ξg)αβ=ξγxγgαβ+gγβxαξγ+gαγxβξγ=0.
Use the commas go to semicolons rule, to obtain
(33.50) ξ γ g α β ; γ + g γ β ξ ; α γ + g α γ ξ ; β γ = 0 . (33.50) ξ γ g α β ; γ + g γ β ξ ; α γ + g α γ ξ ; β γ = 0 . {:(33.50)xi^(gamma)g_(alpha beta;gamma)+g_(gamma beta)xi_(;alpha)^(gamma)+g_(alpha gamma)xi_(;beta)^(gamma)=0.:}\begin{equation*} \xi^{\gamma} g_{\alpha \beta ; \gamma}+g_{\gamma \beta} \xi_{; \alpha}^{\gamma}+g_{\alpha \gamma} \xi_{; \beta}^{\gamma}=0 . \tag{33.50} \end{equation*}(33.50)ξγgαβ;γ+gγβξ;αγ+gαγξ;βγ=0.
Finally, note that for any affine connection 13 g α β ; γ = 0 13 g α β ; γ = 0 ^(13)g_(alpha beta;gamma)=0{ }^{13} g_{\alpha \beta ; \gamma}=013gαβ;γ=0, to obtain
(33.51) ξ α ; β + ξ β ; α = 0 . (33.51) ξ α ; β + ξ β ; α = 0 . {:(33.51)xi_(alpha;beta)+xi_(beta;alpha)=0.:}\begin{equation*} \xi_{\alpha ; \beta}+\xi_{\beta ; \alpha}=0 . \tag{33.51} \end{equation*}(33.51)ξα;β+ξβ;α=0.
Often Killing vectors can be written down by inspection of the form of the metric. From Example 33.6, we see that if ξ ξ xi\boldsymbol{\xi}ξ were to be, say, the basis vector e 1 e 1 e_(1)\boldsymbol{e}_{1}e1, then we have
(33.52) ( £ e 1 g ) α β = x 1 g α β = 0 (33.52) £ e 1 g α β = x 1 g α β = 0 {:(33.52)(£_(e_(1))g)_(alpha beta)=(del)/(delx^(1))g_(alpha beta)=0:}\begin{equation*} \left(£_{e_{1}} \boldsymbol{g}\right)_{\alpha \beta}=\frac{\partial}{\partial x^{1}} g_{\alpha \beta}=0 \tag{33.52} \end{equation*}(33.52)(£e1g)αβ=x1gαβ=0
and so the metric components can't depend on coordinate x 1 x 1 x^(1)x^{1}x1. Therefore, a metric that is independent of x 1 x 1 x^(1)x^{1}x1 has a Killing vector e 1 e 1 e_(1)\boldsymbol{e}_{1}e1. This is the most straightforward way to find Killing vectors: simply see which coordinates do not feature in any of the components of the metric and identify the corresponding basis vectors. Finally, we need to know how to find conserved quantities. Here's the rule:
The dot product u ξ u ξ u*xi\boldsymbol{u} \cdot \boldsymbol{\xi}uξ of a Killing field ξ ξ xi\boldsymbol{\xi}ξ and tangent u u u\boldsymbol{u}u to a geodesic is a constant of the motion along that geodesic.
Example 33.10
This latter claim is easily proved by using the covariant derivative u u grad_(u)\nabla_{u}u to find the change along the geodesic of the dot product u ( u ξ ) u ( u ξ ) grad_(u)(u*xi)\nabla_{\boldsymbol{u}}(\boldsymbol{u} \cdot \boldsymbol{\xi})u(uξ). Using the Leibniz rule, we have
u ( u ξ ) = ( u u ) ξ + u u ξ u ( u ξ ) = u u ξ + u u ξ grad_(u)(u*xi)=(grad_(u)u)*xi+u*grad_(u)xi\boldsymbol{\nabla}_{\boldsymbol{u}}(\boldsymbol{u} \cdot \boldsymbol{\xi})=\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}\right) \cdot \boldsymbol{\xi}+\boldsymbol{u} \cdot \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\xi}u(uξ)=(uu)ξ+uuξ
(33.53)
By definition, a geodesic parallel transports its own tangent vector, meaning u u = u u = grad_(u)u=\nabla_{u} u=uu= 0 , and so the first term is zero. The second term is
(33.54) u u ξ = u α u β ξ α ; β (33.54) u u ξ = u α u β ξ α ; β {:(33.54)u*grad_(u)xi=u^(alpha)u^(beta)xi_(alpha;beta):}\begin{equation*} \boldsymbol{u} \cdot \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\xi}=u^{\alpha} u^{\beta} \xi_{\alpha ; \beta} \tag{33.54} \end{equation*}(33.54)uuξ=uαuβξα;β
Since we're contracting both indices against the components of u u u\boldsymbol{u}u then this last term is symmetric in α α alpha\alphaα and β β beta\betaβ, allowing us to write
(33.55) u ( u ξ ) = u α u β ξ ( α ; β ) = 0 (33.55) u ( u ξ ) = u α u β ξ ( α ; β ) = 0 {:(33.55)grad_(u)(u*xi)=u^(alpha)u^(beta)xi_((alpha;beta))=0:}\begin{equation*} \nabla_{u}(\boldsymbol{u} \cdot \boldsymbol{\xi})=u^{\alpha} u^{\beta} \xi_{(\alpha ; \beta)}=0 \tag{33.55} \end{equation*}(33.55)u(uξ)=uαuβξ(α;β)=0
where the final equality follows from Killing's equation ξ ( α ; β ) = 0 ξ ( α ; β ) = 0 xi_((alpha;beta))=0\xi_{(\alpha ; \beta)}=0ξ(α;β)=0. We conclude that the quantity u ξ u ξ u*xi\boldsymbol{u} \cdot \boldsymbol{\xi}uξ is unchanged along the geodesic.
Let's now use this toolkit to extract some Killing vectors.
Example 33.11
The Schwarzschild metric has a line element
(33.56) d s 2 = ( 1 2 M r ) d t 2 + ( 1 2 M r ) 1 d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) (33.56) d s 2 = 1 2 M r d t 2 + 1 2 M r 1 d r 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 {:(33.56)ds^(2)=(1-(2M)/(r))dt^(2)+(1-(2M)/(r))^(-1)dr^(2)+r^(2)((d)theta^(2)+sin^(2)theta(d)phi^(2)):}\begin{equation*} \mathrm{d} s^{2}=\left(1-\frac{2 M}{r}\right) \mathrm{d} t^{2}+\left(1-\frac{2 M}{r}\right)^{-1} \mathrm{~d} r^{2}+r^{2}\left(\mathrm{~d} \theta^{2}+\sin ^{2} \theta \mathrm{~d} \phi^{2}\right) \tag{33.56} \end{equation*}(33.56)ds2=(12Mr)dt2+(12Mr)1 dr2+r2( dθ2+sin2θ dϕ2)
The components have no dependence on t t ttt and ϕ ϕ phi\phiϕ and so we have Killing vectors
(33.57) e t = t , e ϕ = ϕ (33.57) e t = t , e ϕ = ϕ {:(33.57)e_(t)=(del)/(del t)","quade_(phi)=(del)/(del phi):}\begin{equation*} e_{t}=\frac{\partial}{\partial t}, \quad e_{\phi}=\frac{\partial}{\partial \phi} \tag{33.57} \end{equation*}(33.57)et=t,eϕ=ϕ
As we saw in Chapter 22, this leads to conserved quantities along geodesics of
where u u u\boldsymbol{u}u is tangent to the geodesics.

Example 33.12

Consider Rindler spacetime with metric line element d s 2 = x 2 d t 2 + d x 2 d s 2 = x 2 d t 2 + d x 2 ds^(2)=-x^(2)dt^(2)+dx^(2)\mathrm{d} s^{2}=-x^{2} \mathrm{~d} t^{2}+\mathrm{d} x^{2}ds2=x2 dt2+dx2. Labelling coordinates ( t , x ) ( t , x ) (t,x)(t, x)(t,x), we see that the coordinate t t ttt does not feature in any of the components of the metric. The vector ξ = e t ξ = e t xi=e_(t)\boldsymbol{\xi}=\boldsymbol{e}_{t}ξ=et is therefore the Killing vector. This implies that we have a conserved quantity along the geodesic of u ξ = u t u ξ = u t u*xi=u_(t)\boldsymbol{u} \cdot \boldsymbol{\xi}=u_{t}uξ=ut or
u ξ = g μ ν u μ δ ν t = g t t u t = x 2 u t u ξ = g μ ν u μ δ ν t = g t t u t = x 2 u t u*xi=g_(mu nu)u^(mu)delta_(nu)^(t)=g_(tt)u^(t)=-x^(2)u^(t)\boldsymbol{u} \cdot \boldsymbol{\xi}=g_{\mu \nu} u^{\mu} \delta_{\nu}^{t}=g_{t t} u^{t}=-x^{2} u^{t}uξ=gμνuμδνt=gttut=x2ut

Chapter summary

  • The exterior derivative operator d d d\boldsymbol{d}d inputs a p p ppp-form and outputs a ( p + 1 ) ( p + 1 ) (p+1)(p+1)(p+1)-form.
  • The action of d d d\boldsymbol{d}d on a p p ppp-form A ~ A ~ tilde(A)\tilde{\boldsymbol{A}}A~ is given by
(33.60) d A ~ = 1 p ! A α β σ x μ d x μ d x α d x β d x σ (33.60) d A ~ = 1 p ! A α β σ x μ d x μ d x α d x β d x σ {:(33.60)d tilde(A)=(1)/(p!)*(delA_(alpha beta dots sigma))/(delx^(mu))*dx^(mu)^^dx^(alpha)^^dx^(beta)^^dots^^dx^(sigma):}\begin{equation*} \boldsymbol{d} \tilde{\boldsymbol{A}}=\frac{1}{p!} \cdot \frac{\partial A_{\alpha \beta \ldots \sigma}}{\partial x^{\mu}} \cdot \boldsymbol{d} x^{\mu} \wedge \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \wedge \ldots \wedge \boldsymbol{d} x^{\sigma} \tag{33.60} \end{equation*}(33.60)dA~=1p!Aαβσxμdxμdxαdxβdxσ
  • The Lie derivative is a method of taking derivatives of a tensor field. It requires an additional vector field u u u\boldsymbol{u}u.
  • The Lie derivative of a vector v v v\boldsymbol{v}v with respect to u u u\boldsymbol{u}u is £ u v = [ u , v ] £ u v = [ u , v ] £_(u)v=[u,v]£_{\boldsymbol{u}} \boldsymbol{v}=[\boldsymbol{u}, \boldsymbol{v}]£uv=[u,v].
  • Killing vectors allow us access to conserved quantities in geometrical theories.
  • Along a geodesic with a tangent vector field u u u\boldsymbol{u}u, the quantity u ξ u ξ u*xi\boldsymbol{u} \cdot \boldsymbol{\xi}uξ is conserved, where ξ ξ xi\boldsymbol{\xi}ξ is a Killing vector.

Exercises

(33.1) Consider the amount by which the figure fails to close in Example 33.3. We shall write the distance between initial and final points as
P 4 P 3 = [ u ( P 0 ) + v ( P 1 ) ] [ u ( P 2 ) + v ( P 0 ) ] P 4 P 3 = u P 0 + v P 1 u P 2 + v P 0 P_(4)-P_(3)=[u(P_(0))+v(P_(1))]-[u(P_(2))+v(P_(0))]\mathcal{P}_{4}-\mathcal{P}_{3}=\left[\boldsymbol{u}\left(\mathcal{P}_{0}\right)+\boldsymbol{v}\left(\mathcal{P}_{1}\right)\right]-\left[\boldsymbol{u}\left(\mathcal{P}_{2}\right)+\boldsymbol{v}\left(\mathcal{P}_{0}\right)\right]P4P3=[u(P0)+v(P1)][u(P2)+v(P0)]
where u ( P i ) u P i u(P_(i))\boldsymbol{u}\left(\mathcal{P}_{i}\right)u(Pi) is the vector field u u u\boldsymbol{u}u evaluated at point P i P i P_(i)\mathcal{P}_{i}Pi. Show that this expression yields the commutator of the vector fields u u u\boldsymbol{u}u and v v v\boldsymbol{v}v, evaluated at P 0 P 0 P_(0)\mathcal{P}_{0}P0.
(33.2) (a) Determine the components of the Lie derivative of the ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) tensor A A A\boldsymbol{A}A, written ( £ u A ) ν μ £ u A ν μ (£_(u)A)_(nu)^(mu)\left(£_{u} \boldsymbol{A}\right)_{\nu}^{\mu}(£uA)νμ.
(b) Contraction can be thought of as multiplication by the Kronecker delta δ μ ν = ω μ , e ν δ μ ν = ω μ , e ν delta^(mu)_(nu)=(:omega^(mu),e_(nu):)\delta^{\mu}{ }_{\nu}=\left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{e}_{\nu}\right\rangleδμν=ωμ,eν. By
taking a Lie derivative of δ μ ν δ μ ν delta^(mu)_(nu)\delta^{\mu}{ }_{\nu}δμν, show that Lie differentiation commutes with contraction.
(33.3) Show that
(33.62) £ u £ v £ v £ u = £ [ u , v ] (33.62) £ u £ v £ v £ u = £ [ u , v ] {:(33.62)£_(u)£_(v)-£_(v)£_(u)=£_([u,v]):}\begin{equation*} £_{u} £_{v}-£_{\boldsymbol{v}} £_{\boldsymbol{u}}=£_{[\boldsymbol{u}, \boldsymbol{v}]} \tag{33.62} \end{equation*}(33.62)£u£v£v£u=£[u,v]
by acting with this combination on (i) a scalar function and (ii) a vector field.
Hint: For (ii), use the Jacobi identity [ x , [ y , z ] ] + [ x , [ y , z ] ] + [x,[y,z]]+[\boldsymbol{x},[\boldsymbol{y}, \boldsymbol{z}]]+[x,[y,z]]+ [ z , [ x , y ] ] + [ y , [ z , x ] ] = 0 [ z , [ x , y ] ] + [ y , [ z , x ] ] = 0 [z,[x,y]]+[y,[z,x]]=0[\boldsymbol{z},[\boldsymbol{x}, \boldsymbol{y}]]+[\boldsymbol{y},[\boldsymbol{z}, \boldsymbol{x}]]=0[z,[x,y]]+[y,[z,x]]=0.
(33.4) Using the results of the last exercise, show that the commutator of two Killing fields is a Killing field.

Geometry of the connection

Difficult you call it, Sir? I wish it were impossible. Samuel Johnson (1709-1784) on hearing a famous violinist

To understand the physics of relativity we need the notion of a derivative to encode rates of change. So far in this part of the book, we have encountered two derivatives: the exterior derivative and the Lie derivative. The exterior derivative is designed to work on forms only. The Lie derivative, which works on all tensors, depends on the behaviour of a field at two points, which is unlike the partial derivative in ordinary calculus. Neither of these derivatives is particularly satisfactory to describe the physics of gravitation. In order to have a satisfactory derivative, we need to add a little more structure to our primitive spaces (or manifolds). The solution is found by adding the notion of parallelism to our toolkit. 1 1 ^(1){ }^{1}1 In order to compare two vectors at different points in space, we must have the ability to parallel transport one to the other's location. Being able to keep a vector parallel requires that we must be able to set up basis vectors and 1-forms at each point and, somehow, to connect them. This then allows us to form the covariant derivative that compares a vector at a point to its parallel transported counterpart from another point. It is this derivative that is the most useful for formulating general relativity.
We therefore define 2 2 ^(2){ }^{2}2 a connection, with symbol grad\boldsymbol{\nabla}. This is not a tensor itself, but can be applied to any tensor field and, like the conventional derivative, only takes in the properties of space at a single point. Like the exterior derivative d d d\boldsymbol{d}d, the connection can be thought of as a ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1) object. In fact, the exterior derivative d d d\boldsymbol{d}d and the connection grad\boldsymbol{\nabla} are identical in their action on scalars. 3 3 ^(3){ }^{3}3 As we shall see, the introduction of grad\boldsymbol{\nabla}, which could be viewed as an upgrade of the exterior derivative d d ddd, allows access to an efficient means of extracting the all-important curvature of a spacetime.
The connection can be used to measure the change in a quantity on being transported along a vector u u u\boldsymbol{u}u. The connection directed along u u u\boldsymbol{u}u is u u u*grad\boldsymbol{u} \cdot \boldsymbol{\nabla}u, which is the covariant derivative and given the symbol u u grad_(u)\boldsymbol{\nabla}_{\boldsymbol{u}}u. The covariant derivative tells us to move along the integral curves of the field u u u\boldsymbol{u}u, comparing tensors as we go, using parallel transport to account for any changes in coordinate system at different points in space. Rules for the workings of the connection are given in the margin.
4.1 Covariant derivative in pictures 352
34.2 Connection and exterior derivative and exterior
34.3 Covariant derivative of tensors

1 1 ^(1){ }^{1}1 We shall see in more detail in this chapter how parallelism relies on having a metric. This fundamentally links the covariant derivative and the metric. Since general relativity of the physics of the metric field, this explains why the covariant derivative features so heavily in the description of the mathematics of relativity.
2 2 ^(2){ }^{2}2 The abstract treatment of the covariant derivative u u grad_(u)\boldsymbol{\nabla}_{\boldsymbol{u}}u was developed by Jean-Louis Koszul (1921-2018). The operator grad\boldsymbol{\nabla} wasn't employed in anger until around 1954 in a paper by Katsumi Nomizu (1924-2008) who calls it t t ttt rather than grad\nabla
3 3 ^(3){ }^{3}3 When a connection is present, the action of d d d\boldsymbol{d}d and grad\boldsymbol{\nabla} on vectors, or indeed on any ( n , 0 n , 0 n,0n, 0n,0 ) tensor, will be identical. However, d d d\boldsymbol{d}d and grad\boldsymbol{\nabla} are not identical in their action on forms
In order for the connection to only use the information at a single point, we require linearity in the direction along which we point it:
a u + b v w = a u w + b v w a u + b v w = a u w + b v w grad_(au+bv)w=agrad_(u)w+bgrad_(v)w\boldsymbol{\nabla}_{a \boldsymbol{u}+b \boldsymbol{v}} \boldsymbol{w}=a \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{w}+b \boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{w}au+bvw=auw+bvw. (34.1)
We also want it to be linear in the argument on which it operates
u ( α c + β d ) = α u c + β u d u ( α c + β d ) = α u c + β u d grad_(u)(alpha c+beta d)=alphagrad_(u)c+betagrad_(u)d\nabla_{u}(\alpha c+\beta d)=\alpha \nabla_{u} \boldsymbol{c}+\beta \nabla_{\boldsymbol{u}} \boldsymbol{d}u(αc+βd)=αuc+βud
Finally, we want the connection to obey a Leibniz product rule ( S T ) = ( S T ) + ( S T ) ( S T ) = ( S T ) + ( S T ) grad(S ox T)=(grad S ox T)+(S ox grad T)\boldsymbol{\nabla}(\boldsymbol{S} \otimes \boldsymbol{T})=(\boldsymbol{\nabla} \boldsymbol{S} \otimes \boldsymbol{T})+(\boldsymbol{S} \otimes \boldsymbol{\nabla} \boldsymbol{T})(ST)=(ST)+(ST).
4 4 ^(4){ }^{4}4 The ( 1,2 ) torsion tensor τ τ tau\boldsymbol{\tau}τ along with the ( 1,3 ) Riemann curvature tensor R R R\boldsymbol{R}R are two independent characteristic features of a connection. The torsion-free ( τ = 0 ) ( τ = 0 ) (tau=0)(\boldsymbol{\tau}=0)(τ=0) spaces we consider in general relativity all share the property of symmetry of the connection
Γ α β μ = Γ β α μ , Γ α β μ = Γ β α μ , Gamma_(alpha beta)^(mu)=Gamma_(beta alpha)^(mu),\Gamma_{\alpha \beta}^{\mu}=\Gamma_{\beta \alpha}^{\mu},Γαβμ=Γβαμ,
when expressed in a coordinate basis.
Fig. 34.1 After the vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v have been transported, the quadrilateral fails to close.
5 5 ^(5){ }^{5}5 The symbol u v u v grad_(u)v\nabla_{u} vuv tells us to transport the vector v v v\boldsymbol{v}v along the vector field u u u\boldsymbol{u}u and measure the difference in v v v\boldsymbol{v}v. We could, of course, transport the vector field u u u\boldsymbol{u}u along the vector v v v\boldsymbol{v}v by considering v u v u grad_(v)u\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}vu.

34.1 Covariant derivative in pictures

In the sorts of manifolds with which we work in general relativity, there is a neat way of visualizing the properties of the covariant derivative.
Although general relativity embraces a geometrical description of gravity, it does so by singling out curvature as the fundamental geometrical object associated with a connection, to the detriment of a quantity called torsion. The torsion of two vector fields u u u\boldsymbol{u}u and v v v\boldsymbol{v}v is a ( 1,2 ) tensor τ τ tau\tauτ defined by
(34.4) τ ( u , v ) = u v v u [ u , v ] . (34.4) τ ( u , v ) = u v v u [ u , v ] . {:(34.4)tau(u","v)=grad_(u)v-grad_(v)u-[u","v].:}\begin{equation*} \boldsymbol{\tau}(\boldsymbol{u}, \boldsymbol{v})=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}-[\boldsymbol{u}, \boldsymbol{v}] . \tag{34.4} \end{equation*}(34.4)τ(u,v)=uvvu[u,v].
It's certainly worth noting that alternative theories of gravity can be formulated in terms of torsion, which potentially provide a richer phenomenology than the conventional curvature-based theory. However, in the spaces that we conventionally consider in general relativity, the torsion vanishes, 4 4 ^(4){ }^{4}4 so we always have the property that
(34.5) u v v u = [ u , v ] (34.5) u v v u = [ u , v ] {:(34.5)grad_(u)v-grad_(v)u=[u","v]:}\begin{equation*} \boldsymbol{\nabla}_{u} v-\boldsymbol{\nabla}_{v} u=[\boldsymbol{u}, \boldsymbol{v}] \tag{34.5} \end{equation*}(34.5)uvvu=[u,v]
As we see in the next example, this condition allows us to relate the covariant and Lie derivatives.
Example 34.1
When the torsion vanishes, the covariant derivatives of vectors are related to Lie derivatives via
£ u v = [ u , v ] = u v v u , (34.6) ( £ u v ) α = v α ; β u β u α ; β v β , £ u v = [ u , v ] = u v v u , (34.6) £ u v α = v α ; β u β u α ; β v β , {:[£_(u)v=[u","v]=grad_(u)v-grad_(v)u","],[(34.6)(£_(u)v)^(alpha)=v^(alpha)_(;beta)u^(beta)-u^(alpha)_(;beta)v^(beta)","]:}\begin{align*} £_{\boldsymbol{u}} \boldsymbol{v} & =[\boldsymbol{u}, \boldsymbol{v}]=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}, \\ \left(£_{\boldsymbol{u}} \boldsymbol{v}\right)^{\alpha} & =v^{\alpha}{ }_{; \beta} u^{\beta}-u^{\alpha}{ }_{; \beta} v^{\beta}, \tag{34.6} \end{align*}£uv=[u,v]=uvvu,(34.6)(£uv)α=vα;βuβuα;βvβ,
where the second expression is given in terms of components and semicolon notation.
Since the vanishing of torsion allows us to link the connection and the Lie derivative, we can also interpret the connection visually, in terms of whether a figure fails to close, in the same way that we interpreted the Lie derivative in the last chapter.

Example 34.2

As in the last chapter, we evaluate the vector that closes the figure v v v\boldsymbol{v}v - u v u u v u u-v-u\boldsymbol{u}-\boldsymbol{v}-\boldsymbol{u}uvu. As we've seen, this is given by the commutator | v , u | = v u u v | v , u | = v u u v |v,u|=vu-uv|\boldsymbol{v}, \boldsymbol{u}|=\boldsymbol{v} \boldsymbol{u}-\boldsymbol{u} \boldsymbol{v}|v,u|=vuuv. Consider Fig. 34.1, where we use the parameters λ λ lambda\lambdaλ and σ σ sigma\sigmaσ to parametrize the vector fields v v v\boldsymbol{v}v and u u u\boldsymbol{u}u respectively. The figure is formed from the vector field v v v\boldsymbol{v}v evaluated at parameter point λ 0 λ 0 lambda_(0)\lambda_{0}λ0 and λ 0 + 1 λ 0 + 1 lambda_(0)+1\lambda_{0}+1λ0+1, which gives us two different vectors. (They are different vectors, because we've evaluated the vector field at two different points.) We also evaluate the field we've evaluated the vector field at two different points.
u u u\boldsymbol{u}u at points σ 0 σ 0 sigma_(0)\sigma_{0}σ0 and σ 0 + 1 σ 0 + 1 sigma_(0)+1\sigma_{0}+1σ0+1 and use these four vectors to form the figure. We see that u u u\boldsymbol{u}u at points σ 0 σ 0 sigma_(0)\sigma_{0}σ0 and σ 0 + 1 σ 0 + 1 sigma_(0)+1\sigma_{0}+1σ0+1 and use these four ve
the figure does not close. Using the fact that 5 5 ^(5){ }^{5}5
v ( λ 0 + 1 ) v ( λ 0 ) u v , (34.7) u ( σ 0 + 1 ) u ( σ 0 ) v u v λ 0 + 1 v λ 0 u v , (34.7) u σ 0 + 1 u σ 0 v u {:[v(lambda_(0)+1)-v(lambda_(0))~~grad_(u)v","],[(34.7)u(sigma_(0)+1)-u(sigma_(0))~~grad_(v)u]:}\begin{align*} & v\left(\lambda_{0}+1\right)-v\left(\lambda_{0}\right) \approx \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}, \\ & u\left(\sigma_{0}+1\right)-u\left(\sigma_{0}\right) \approx \boldsymbol{\nabla}_{v} \boldsymbol{u} \tag{34.7} \end{align*}v(λ0+1)v(λ0)uv,(34.7)u(σ0+1)u(σ0)vu
we see that the amount by which it fails to close is given by u v v u u v v u grad_(u)v-grad_(v)u\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}uvvu. It follows that u v v u = [ u , v ] u v v u = [ u , v ] grad_(u)v-grad_(v)u=[u,v]\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{u}=[\boldsymbol{u}, \boldsymbol{v}]uvvu=[u,v].
For cases where the commutator [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v] vanishes (i.e. when the quadrilateral closes), we have the useful property of the symmetry of the covariant derivative which says
(34.8) u v = v u (34.8) u v = v u {:(34.8)grad_(u)v=grad_(v)u:}\begin{equation*} \nabla_{u} v=\nabla_{v} u \tag{34.8} \end{equation*}(34.8)uv=vu

34.2 Connection and exterior derivative

How do we understand the connection symbol grad\boldsymbol{\nabla} ? That is to say, what is the meaning of a covariant derivative without the specification of a direction along which to take the derivative? The connection grad\boldsymbol{\nabla} is much like the exterior derivative d d d\boldsymbol{d}d, and is defined on a scalar function via
(34.9) f = d f (34.9) f = d f {:(34.9)grad f=df:}\begin{equation*} \nabla f=\boldsymbol{d} f \tag{34.9} \end{equation*}(34.9)f=df
This some formal section introduces that will  This   some formal section introduces that will  ↷" This "_(" some formal section introduces that will ")\underset{\text { some formal section introduces that will }}{\curvearrowright \text { This }} This  some formal section introduces that will  some formal ideas that will be taken up in Chapter 36 to formulate a highly effient method for extract
ture from a metric.
The resulting quantity is a 1 -form. We can give it a direction using d f , u = u f d f , u = u f (:df,u:)=del_(u)f\langle\boldsymbol{d} f, \boldsymbol{u}\rangle=\partial_{\boldsymbol{u}} fdf,u=uf, or equivalently
(34.10) u f = u f u μ f x μ u f (34.10) u f = u f u μ f x μ u f {:(34.10)u*grad f=grad_(u)f-=u^(mu)(del f)/(delx^(mu))-=del_(u)f:}\begin{equation*} \boldsymbol{u} \cdot \nabla f=\nabla_{u} f \equiv u^{\mu} \frac{\partial f}{\partial x^{\mu}} \equiv \partial_{\boldsymbol{u}} f \tag{34.10} \end{equation*}(34.10)uf=ufuμfxμuf
where we've written the results in several forms of notation. 6 6 ^(6){ }^{6}6
Previously, we confined the use of the differential operator d d d\boldsymbol{d}d to forms [i.e. antisymmetric ( 0 , n ) ( 0 , n ) (0,n)(0, n)(0,n) objects] and, in this spirit, we used it above on a function, which is a 0 -form. We now extend the use of this tool to vectors, which are ( 1 , 0 ) ( 1 , 0 ) (1,0)(1,0)(1,0) objects. When acting on vectors, the action of d d d\boldsymbol{d}d and grad\boldsymbol{\nabla} are identical. We interpret the action of d d d\boldsymbol{d}d and grad\boldsymbol{\nabla} on a vector as saying that the object is vector-valued [that is, there's a 1 in the first slot of the valency ( 1 , 0 ) ( 1 , 0 ) (1,0)(1,0)(1,0) ], but that it is also a 0 -form [that is, the 0 part of ( 1 , 0 ) ( 1 , 0 ) (1,0)(1,0)(1,0) ]. Crucial here is having a connection available. In its absence, the action of d d d\boldsymbol{d}d on a vector is undefined.
In component notation, the vector v v v\boldsymbol{v}v is written as v μ e μ v μ e μ v^(mu)e_(mu)v^{\mu} \boldsymbol{e}_{\mu}vμeμ. Acting on this with d d d\boldsymbol{d}d gives rise to (i) a contribution from d v μ d v μ dv^(mu)\boldsymbol{d} v^{\mu}dvμ, which is simply the action of d d d\boldsymbol{d}d on a scalar function, and (ii) a contribution from the action on the vector-valued 0 -form e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ. The covariant derivative of a vector v = v μ e μ v = v μ e μ v=v^(mu)e_(mu)\boldsymbol{v}=v^{\mu} \boldsymbol{e}_{\mu}v=vμeμ can then be written as
(34.12) v d v = d v μ e μ + v μ d e μ (34.12) v d v = d v μ e μ + v μ d e μ {:(34.12)grad v-=dv=dv^(mu)oxe_(mu)+v^(mu)de_(mu):}\begin{equation*} \nabla \boldsymbol{v} \equiv \boldsymbol{d} \boldsymbol{v}=\boldsymbol{d} v^{\mu} \otimes \boldsymbol{e}_{\mu}+v^{\mu} \boldsymbol{d} \boldsymbol{e}_{\mu} \tag{34.12} \end{equation*}(34.12)vdv=dvμeμ+vμdeμ
The quantity d v d v dv\boldsymbol{d} \boldsymbol{v}dv is known as a vector-valued 1-form. 7 7 ^(7){ }^{7}7 Note that it depends on d e μ d e μ de_(mu)d e_{\mu}deμ, which tells us how the basis vectors change in space. This quantity gives the nature of the connection. We therefore define 8 8 ^(8){ }^{8}8 connection coefficients in terms of the action of d ( d ( d(\boldsymbol{d}(d( or ) ) grad)\boldsymbol{\nabla})) on the basis vectors e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ as follows:
(34.14) e μ d e μ = Γ ν μ α ( ω ν e α ) . (34.14) e μ d e μ = Γ ν μ α ω ν e α . {:(34.14)grade_(mu)-=de_(mu)=Gamma_(nu mu)^(alpha)(omega^(nu)oxe_(alpha)).:}\begin{equation*} \nabla \boldsymbol{e}_{\mu} \equiv \boldsymbol{d} \boldsymbol{e}_{\mu}=\Gamma_{\nu \mu}^{\alpha}\left(\boldsymbol{\omega}^{\nu} \otimes \boldsymbol{e}_{\alpha}\right) . \tag{34.14} \end{equation*}(34.14)eμdeμ=Γνμα(ωνeα).
We can relate this more geometric definition of the connection to our previous notion of connection coefficients (see eqn 7.6) in the following example.
6 6 ^(6){ }^{6}6 This doesn't exhaust the notational possibilities. For example, an equivalent description in a more geometrical notation is
(34.11) u f d f , u u [ f ] (34.11) u f d f , u u [ f ] {:(34.11)grad_(u)f-=(:df","u:)-=u[f]:}\begin{equation*} \boldsymbol{\nabla}_{\boldsymbol{u}} f \equiv\langle\boldsymbol{d} f, \boldsymbol{u}\rangle \equiv \boldsymbol{u}[f] \tag{34.11} \end{equation*}(34.11)ufdf,uu[f]
7 7 ^(7){ }^{7}7 Why not simply call this a ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) tensor? It's more subtle than that. A general ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) tensor may be written as
(34.13) S = S ν μ ω ν e μ (34.13) S = S ν μ ω ν e μ {:(34.13)S=S_(nu)^(mu)omega^(nu)oxe_(mu):}\begin{equation*} \boldsymbol{S}=S_{\nu}^{\mu} \boldsymbol{\omega}^{\nu} \otimes \boldsymbol{e}_{\mu} \tag{34.13} \end{equation*}(34.13)S=Sνμωνeμ
However, the vector-valued 1 -form features the term d e μ d e μ de_(mu)d e_{\mu}deμ, which contains information about how the basis vectors themselves change in space. This relies on the connection, and so the vectorvalued 1 -form requires more structure than is needed to define a standard tensor like S S S\boldsymbol{S}S.
8 8 ^(8){ }^{8}8 Previously (Chapter 9), we found the connection coefficients by considering derivatives of the metric, which has been notable by its absence in this part of the book. We shall see at the end of the chapter how parallelism, that gives us the covariant derivative (and therefore these coefficients), relies on having a metric.

Example 34.3

Noting that A , e β = β A A , e β = β A (:grad A,e_(beta):)=grad_(beta)A\left\langle\boldsymbol{\nabla} \boldsymbol{A}, \boldsymbol{e}_{\beta}\right\rangle=\boldsymbol{\nabla}_{\beta} \boldsymbol{A}A,eβ=βA, we have
β e μ = Γ α ν μ ( ω ν e α ) , e β β e μ = Γ α ν μ ω ν e α , e β grad_(beta)e_(mu)=Gamma^(alpha)_(nu mu)(:(omega^(nu)oxe_(alpha)),e_(beta):)\boldsymbol{\nabla}_{\beta} \boldsymbol{e}_{\mu}=\Gamma^{\alpha}{ }_{\nu \mu}\left\langle\left(\boldsymbol{\omega}^{\nu} \otimes \boldsymbol{e}_{\alpha}\right), \boldsymbol{e}_{\beta}\right\rangleβeμ=Γανμ(ωνeα),eβ = Γ α ν μ ω ν ( e β ) e α = Γ α ν μ ω ν e β e α =Gamma^(alpha)_(nu mu)omega^(nu)(e_(beta))oxe_(alpha)=\Gamma^{\alpha}{ }_{\nu \mu} \boldsymbol{\omega}^{\nu}\left(\boldsymbol{e}_{\beta}\right) \otimes \boldsymbol{e}_{\alpha}=Γανμων(eβ)eα = Γ α ν μ δ ν β e α = Γ α β μ e α = Γ α ν μ δ ν β e α = Γ α β μ e α =Gamma^(alpha)_(nu mu)delta^(nu)_(beta)e_(alpha)=Gamma^(alpha)_(beta mu)e_(alpha)quad=\Gamma^{\alpha}{ }_{\nu \mu} \delta^{\nu}{ }_{\beta} \boldsymbol{e}_{\alpha}=\Gamma^{\alpha}{ }_{\beta \mu} \boldsymbol{e}_{\alpha} \quad=Γανμδνβeα=Γαβμeα (using ω ν ( e β ) = δ ν β ω ν e β = δ ν β omega^(nu)(e_(beta))=delta^(nu)_(beta)\boldsymbol{\omega}^{\nu}\left(\boldsymbol{e}_{\beta}\right)=\delta^{\nu}{ }_{\beta}ων(eβ)=δνβ ),
or, picking out components by taking an inner product with ω σ ω σ omega^(sigma)\boldsymbol{\omega}^{\sigma}ωσ
(34.15) ω σ , β e μ = Γ β μ σ . (34.15) ω σ , β e μ = Γ β μ σ . {:(34.15)(:omega^(sigma),grad_(beta)e_(mu):)=Gamma_(beta mu)^(sigma).:}\begin{equation*} \left\langle\boldsymbol{\omega}^{\sigma}, \boldsymbol{\nabla}_{\beta} \boldsymbol{e}_{\mu}\right\rangle=\Gamma_{\beta \mu}^{\sigma} . \tag{34.15} \end{equation*}(34.15)ωσ,βeμ=Γβμσ.
Without worrying about the direction of the derivative, we're now able to write down the connection's action on the vector field v v v\boldsymbol{v}v.

Example 34.4

We can write
v = v μ e μ + v μ e μ (chain rule) (34.16) = d v μ e μ + v α Γ μ β α ( ω β e μ ) (eqn 34.14) = v μ x β ( d x β e μ ) + v α Γ β α μ ( ω β e μ ) (action of d on v μ ) . v = v μ e μ + v μ e μ  (chain rule)  (34.16) = d v μ e μ + v α Γ μ β α ω β e μ  (eqn 34.14)  = v μ x β d x β e μ + v α Γ β α μ ω β e μ  (action of  d  on  v μ . {:[grad v=gradv^(mu)oxe_(mu)+v^(mu)grade_(mu)" (chain rule) "],[(34.16)=dv^(mu)oxe_(mu)+v^(alpha)Gamma^(mu)_(beta alpha)(omega^(beta)oxe_(mu))" (eqn 34.14) "],[=(delv^(mu))/(delx^(beta))(dx^(beta)oxe_(mu))+v^(alpha)Gamma_(beta alpha)^(mu)(omega^(beta)oxe_(mu))" (action of "{:d" on "v_(mu)).]:}\begin{align*} \boldsymbol{\nabla} \boldsymbol{v} & =\boldsymbol{\nabla} v^{\mu} \otimes \boldsymbol{e}_{\mu}+v^{\mu} \boldsymbol{\nabla} \boldsymbol{e}_{\mu} & & \text { (chain rule) } \\ & =\boldsymbol{d} v^{\mu} \otimes \boldsymbol{e}_{\mu}+v^{\alpha} \Gamma^{\mu}{ }_{\beta \alpha}\left(\boldsymbol{\omega}^{\beta} \otimes \boldsymbol{e}_{\mu}\right) & & \text { (eqn 34.14) } \tag{34.16}\\ & =\frac{\partial v^{\mu}}{\partial x^{\beta}}\left(\boldsymbol{d} x^{\beta} \otimes \boldsymbol{e}_{\mu}\right)+v^{\alpha} \Gamma_{\beta \alpha}^{\mu}\left(\boldsymbol{\omega}^{\beta} \otimes \boldsymbol{e}_{\mu}\right) & & \text { (action of } \left.\boldsymbol{d} \text { on } v_{\mu}\right) . \end{align*}v=vμeμ+vμeμ (chain rule) (34.16)=dvμeμ+vαΓμβα(ωβeμ) (eqn 34.14) =vμxβ(dxβeμ)+vαΓβαμ(ωβeμ) (action of d on vμ).
Using the last example and noting that d x β ω β d x β ω β dx^(beta)-=omega^(beta)\boldsymbol{d} x^{\beta} \equiv \boldsymbol{\omega}^{\beta}dxβωβ, we have the final expression,
(34.17) v = ( v μ x β + v α Γ β α μ ) ( ω β e μ ) . (34.17) v = v μ x β + v α Γ β α μ ω β e μ . {:(34.17)grad v=((delv^(mu))/(delx^(beta))+v^(alpha)Gamma_(beta alpha)^(mu))(omega^(beta)oxe_(mu)).:}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{v}=\left(\frac{\partial v^{\mu}}{\partial x^{\beta}}+v^{\alpha} \Gamma_{\beta \alpha}^{\mu}\right)\left(\boldsymbol{\omega}^{\beta} \otimes \boldsymbol{e}_{\mu}\right) . \tag{34.17} \end{equation*}(34.17)v=(vμxβ+vαΓβαμ)(ωβeμ).
Note that this can also be written in semicolon notation as v = v = grad v=\nabla \boldsymbol{v}=v= v μ ; β ( ω β e μ ) v μ ; β ω β e μ v^(mu)_(;beta)(omega^(beta)oxe_(mu))v^{\mu}{ }_{; \beta}\left(\boldsymbol{\omega}^{\beta} \otimes \boldsymbol{e}_{\mu}\right)vμ;β(ωβeμ). That is to say that the components of v v grad v\boldsymbol{\nabla} \boldsymbol{v}v are the covariant derivative components v μ ; β v μ ; β v^(mu)_(;beta)v^{\mu}{ }_{; \beta}vμ;β. We conclude that from the linear slot machine point of view, the connection symbol grad\boldsymbol{\nabla} acting on a vector v v v\boldsymbol{v}v results in the 2-slotted object
(34.18) v ( , ) = v ; ν μ ω ν ( ) e μ ( ) . (34.18) v ( , ) = v ; ν μ ω ν ( ) e μ ( ) . {:(34.18)grad v(",")=v_(;nu)^(mu)omega^(nu)()oxe_(mu)().:}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{v}(,)=v_{; \nu}^{\mu} \boldsymbol{\omega}^{\nu}() \otimes \boldsymbol{e}_{\mu}() . \tag{34.18} \end{equation*}(34.18)v(,)=v;νμων()eμ().
This is the gradient of v v v\boldsymbol{v}v. We summarize the properties of the gradient in the next example.

Example 34.5

The gradient of v v v\boldsymbol{v}v is neither a vector nor a 1-form, but a vector-valued 1-form, sometimes written as d v d v dv\boldsymbol{d} \boldsymbol{v}dv. You must insert both a vector and a 1 -form to get a number. It has components
(34.19) v ( e α , ω μ ) = ( v ) α μ = ( v μ x α + v λ Γ α λ μ ) . (34.19) v e α , ω μ = ( v ) α μ = v μ x α + v λ Γ α λ μ . {:(34.19)grad v(e_(alpha),omega^(mu))=(grad v)_(alpha)^(mu)=((delv^(mu))/(delx^(alpha))+v^(lambda)Gamma_(alpha lambda)^(mu)).:}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{v}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{\omega}^{\mu}\right)=(\boldsymbol{\nabla} \boldsymbol{v})_{\alpha}^{\mu}=\left(\frac{\partial v^{\mu}}{\partial x^{\alpha}}+v^{\lambda} \Gamma_{\alpha \lambda}^{\mu}\right) . \tag{34.19} \end{equation*}(34.19)v(eα,ωμ)=(v)αμ=(vμxα+vλΓαλμ).
If we insert v = e γ v = e γ v=e_(gamma)v=e_{\gamma}v=eγ into this last equation we effectively extract the components of grad\boldsymbol{\nabla}
(34.20) ( e γ ) μ α = δ λ γ Γ μ α λ = Γ α γ μ . (34.20) e γ μ α = δ λ γ Γ μ α λ = Γ α γ μ . {:(34.20)(grade_(gamma))^(mu)_(alpha)=delta^(lambda)_(gamma)Gamma^(mu)_(alpha lambda)=Gamma_(alpha gamma)^(mu).:}\begin{equation*} \left(\boldsymbol{\nabla} \boldsymbol{e}_{\gamma}\right)^{\mu}{ }_{\alpha}=\delta^{\lambda}{ }_{\gamma} \Gamma^{\mu}{ }_{\alpha \lambda}=\Gamma_{\alpha \gamma}^{\mu} . \tag{34.20} \end{equation*}(34.20)(eγ)μα=δλγΓμαλ=Γαγμ.
That is, the components of the connection grad\boldsymbol{\nabla} can be thought of as the connection coefficients.
The quantity v ( u ) = , u v v ( u ) = , u v grad v(u)=,grad_(u)v\boldsymbol{\nabla} \boldsymbol{v}(\boldsymbol{u})=,\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}v(u)=,uv is a vector. It is the covariant derivative of v v v\boldsymbol{v}v along u u uuu with components
(34.21) ω μ , u v = ( u v ) μ = u α ( v μ x α + v λ Γ α λ μ ) (34.21) ω μ , u v = u v μ = u α v μ x α + v λ Γ α λ μ {:(34.21)(:omega^(mu),grad_(u)v:)=(grad_(u)v)^(mu)=u^(alpha)((delv^(mu))/(delx^(alpha))+v^(lambda)Gamma_(alpha lambda)^(mu)):}\begin{equation*} \left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangle=\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right)^{\mu}=u^{\alpha}\left(\frac{\partial v^{\mu}}{\partial x^{\alpha}}+v^{\lambda} \Gamma_{\alpha \lambda}^{\mu}\right) \tag{34.21} \end{equation*}(34.21)ωμ,uv=(uv)μ=uα(vμxα+vλΓαλμ)
To get a number from v v grad v\boldsymbol{\nabla} \boldsymbol{v}v, insert a 1-form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ and a vector u u u\boldsymbol{u}u :
(34.22) v ( u , σ ~ ) = σ ~ , u v = (number) (34.22) v ( u , σ ~ ) = σ ~ , u v =  (number)  {:(34.22)grad v(u"," tilde(sigma))=(:( tilde(sigma)),grad_(u)v:)=" (number) ":}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{v}(\boldsymbol{u}, \tilde{\boldsymbol{\sigma}})=\left\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}\right\rangle=\text { (number) } \tag{34.22} \end{equation*}(34.22)v(u,σ~)=σ~,uv= (number) 
The number that is outputted evaluates the number of times the vector u v u v grad_(u)v\nabla_{u} vuv pierces the surfaces of the 1 -form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~.
Finally, we shall derive the transformation properties of the connection coefficients.
Example 34.6
Relating primed and unprimed frames, we write
Γ β γ α = ω α , e β e γ (34.23) = Λ α α α ω α , Λ β β e β ( Λ γ γ e γ ) . Γ β γ α = ω α , e β e γ (34.23) = Λ α α α ω α , Λ β β e β Λ γ γ e γ . {:[Gamma_(beta^(')gamma^('))^(alpha^('))=(:omega^(alpha^(')),grad_(e_(beta^(')))e_(gamma^(')):)],[(34.23)=(:Lambda_(alpha)^(alpha_(alpha)^('))omega^(alpha),grad_({:Lambda^(beta)_(beta^('))e_(beta)(Lambda_(gamma^('))^(gamma)e_(gamma)):)).:}]:}\begin{align*} \Gamma_{\beta^{\prime} \gamma^{\prime}}^{\alpha^{\prime}} & =\left\langle\boldsymbol{\omega}^{\alpha^{\prime}}, \boldsymbol{\nabla}_{\boldsymbol{e}_{\beta^{\prime}}} \boldsymbol{e}_{\gamma^{\prime}}\right\rangle \\ & =\left\langle\Lambda_{\alpha}^{\alpha_{\alpha}^{\prime}} \boldsymbol{\omega}^{\alpha}, \boldsymbol{\nabla}_{\left.\Lambda^{\beta}{ }_{\beta^{\prime}} \boldsymbol{e}_{\beta}\left(\Lambda_{\gamma^{\prime}}^{\gamma} \boldsymbol{e}_{\gamma}\right)\right\rangle} .\right. \tag{34.23} \end{align*}Γβγα=ωα,eβeγ(34.23)=Λαααωα,Λββeβ(Λγγeγ).
Using linearity and the rule that a u = a u a u = a u grad_(au)=agrad_(u)\boldsymbol{\nabla}_{a \boldsymbol{u}}=a \boldsymbol{\nabla}_{\boldsymbol{u}}au=au we have
(34.24) Γ β γ α = Λ α α Λ β β ω α , e β ( Λ γ γ e γ ) (34.24) Γ β γ α = Λ α α Λ β β ω α , e β Λ γ γ e γ {:(34.24)Gamma_(beta^(')gamma^('))^(alpha^('))=Lambda_(alpha^('))^(alpha^('))Lambda_(beta^('))^(beta)(:omega^(alpha),grad_(e_(beta))(Lambda_(gamma^('))^(gamma)e_(gamma)):):}\begin{equation*} \Gamma_{\beta^{\prime} \gamma^{\prime}}^{\alpha^{\prime}}=\Lambda_{\alpha^{\prime}}^{\alpha^{\prime}} \Lambda_{\beta^{\prime}}^{\beta}\left\langle\boldsymbol{\omega}^{\alpha}, \nabla_{\boldsymbol{e}_{\beta}}\left(\Lambda_{\gamma^{\prime}}^{\gamma} \boldsymbol{e}_{\gamma}\right)\right\rangle \tag{34.24} \end{equation*}(34.24)Γβγα=ΛααΛββωα,eβ(Λγγeγ)
Now using the Leibniz rule and the rule that 9 e 9 e ^(9)grad_(e){ }^{9} \boldsymbol{\nabla}_{\boldsymbol{e}}9e f = e β [ f ] f = e β [ f ] f=e_(beta)[f]f=\boldsymbol{e}_{\beta}[f]f=eβ[f] that we find
Γ α β γ = Λ α α Λ β β ( ω α , e β [ Λ γ γ ] e γ + Λ γ γ Γ α β γ ) = Λ α α Λ β β β e β [ Λ γ γ ] δ α γ + Λ α Λ α β β Λ γ γ Γ α β γ (34.25) = Λ α α e β [ Λ α γ ] + Λ α α Λ β Λ γ γ Γ α β γ . Γ α β γ = Λ α α Λ β β ω α , e β Λ γ γ e γ + Λ γ γ Γ α β γ = Λ α α Λ β β β e β Λ γ γ δ α γ + Λ α Λ α β β Λ γ γ Γ α β γ (34.25) = Λ α α e β Λ α γ + Λ α α Λ β Λ γ γ Γ α β γ . {:[Gamma^(alpha^('))_(beta^(')gamma^('))=Lambda^(alpha_(alpha)^('))Lambda^(beta)_(beta^('))((:omega^(alpha),e_(beta)[Lambda^(gamma)_(gamma^('))]e_(gamma):)+Lambda^(gamma)_(gamma^('))Gamma^(alpha)_(beta gamma))],[=Lambda^(alpha_(alpha)^('))Lambda^(beta)^(beta)_(beta^('))e_(beta)[Lambda^(gamma)_(gamma^('))]delta^(alpha)_(gamma)+Lambda^(alpha^('))Lambda_(alpha)^(beta)_(beta^('))Lambda^(gamma)_(gamma^('))Gamma^(alpha)_(beta gamma)],[(34.25)=Lambda^(alpha_(alpha)^('))e_(beta^('))[Lambda^(alpha)_(gamma^('))]+Lambda^(alpha_(alpha)^('))Lambda_(beta^('))Lambda^(gamma)_(gamma^('))Gamma^(alpha)_(beta gamma).]:}\begin{align*} \Gamma^{\alpha^{\prime}}{ }_{\beta^{\prime} \gamma^{\prime}} & =\Lambda^{\alpha_{\alpha}^{\prime}} \Lambda^{\beta}{ }_{\beta^{\prime}}\left(\left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}\left[\Lambda^{\gamma}{ }_{\gamma^{\prime}}\right] \boldsymbol{e}_{\gamma}\right\rangle+\Lambda^{\gamma}{ }_{\gamma^{\prime}} \Gamma^{\alpha}{ }_{\beta \gamma}\right) \\ & =\Lambda^{\alpha_{\alpha}^{\prime}} \Lambda^{\beta}{ }^{\beta}{ }_{\beta^{\prime}} \boldsymbol{e}_{\beta}\left[\Lambda^{\gamma}{ }_{\gamma^{\prime}}\right] \delta^{\alpha}{ }_{\gamma}+\Lambda^{\alpha^{\prime}} \Lambda_{\alpha}^{\beta}{ }_{\beta^{\prime}} \Lambda^{\gamma}{ }_{\gamma^{\prime}} \Gamma^{\alpha}{ }_{\beta \gamma} \\ & =\Lambda^{\alpha_{\alpha}^{\prime}} \boldsymbol{e}_{\beta^{\prime}}\left[\Lambda^{\alpha}{ }_{\gamma^{\prime}}\right]+\Lambda^{\alpha_{\alpha}^{\prime}} \Lambda_{\beta^{\prime}} \Lambda^{\gamma}{ }_{\gamma^{\prime}} \Gamma^{\alpha}{ }_{\beta \gamma} . \tag{34.25} \end{align*}Γαβγ=ΛααΛββ(ωα,eβ[Λγγ]eγ+ΛγγΓαβγ)=ΛααΛβββeβ[Λγγ]δαγ+ΛαΛαββΛγγΓαβγ(34.25)=Λααeβ[Λαγ]+ΛααΛβΛγγΓαβγ.
The second term is the usual tensor transformation law for the components of a tensor. However, it's the first term that ruins the tensor transformation properties. We can see how if we use a coordinate frame (and recall that Λ α β = x α / x β Λ α β = x α / x β Lambda^(alpha)_(beta)=delx^(alpha)//delx^(beta)\Lambda^{\alpha}{ }_{\beta}=\partial x^{\alpha} / \partial x^{\beta}Λαβ=xα/xβ and e β [ f ] = f / x β ) e β [ f ] = f / x β {:e_(beta)[f]=del f//delx^(beta))\left.\boldsymbol{e}_{\beta}[f]=\partial f / \partial x^{\beta}\right)eβ[f]=f/xβ), where we find
(34.26) Γ β γ α = x α x α 2 x α x β x γ + x α x α x β x β x γ x γ Γ α β γ (34.26) Γ β γ α = x α x α 2 x α x β x γ + x α x α x β x β x γ x γ Γ α β γ {:(34.26)Gamma_(beta^(')gamma^('))^(alpha^('))=(delx^(alpha^(')))/(delx^(alpha))*(del^(2)x^(alpha))/(delx^(beta^('))delx^(gamma^(')))+(delx^(alpha^(')))/(delx^(alpha))(delx^(beta))/(delx^(beta^(')))*(delx^(gamma))/(delx^(gamma^(')))Gamma^(alpha)_(beta gamma):}\begin{equation*} \Gamma_{\beta^{\prime} \gamma^{\prime}}^{\alpha^{\prime}}=\frac{\partial x^{\alpha^{\prime}}}{\partial x^{\alpha}} \cdot \frac{\partial^{2} x^{\alpha}}{\partial x^{\beta^{\prime}} \partial x^{\gamma^{\prime}}}+\frac{\partial x^{\alpha^{\prime}}}{\partial x^{\alpha}} \frac{\partial x^{\beta}}{\partial x^{\beta^{\prime}}} \cdot \frac{\partial x^{\gamma}}{\partial x^{\gamma^{\prime}}} \Gamma^{\alpha}{ }_{\beta \gamma} \tag{34.26} \end{equation*}(34.26)Γβγα=xαxα2xαxβxγ+xαxαxβxβxγxγΓαβγ

34.3 Covariant derivative of tensors

The rule for ( 1 , 0 ) ( 1 , 0 ) (1,0)(1,0)(1,0) objects that d v v d v v dv-=grad v\boldsymbol{d} \boldsymbol{v} \equiv \boldsymbol{\nabla} \boldsymbol{v}dvv carries over to all ( n , 0 ) ( n , 0 ) (n,0)(n, 0)(n,0) tensors, that is, all tensor valued 0 -forms, where we have
(34.27) d S = S (34.27) d S = S {:(34.27)dS=grad S:}\begin{equation*} d S=\nabla S \tag{34.27} \end{equation*}(34.27)dS=S
where S S S\boldsymbol{S}S is a ( n , 0 ) ( n , 0 ) (n,0)(n, 0)(n,0) tensor. It might seem that we can simply make d d d\boldsymbol{d}d equivalent to grad\boldsymbol{\nabla} in its action on all objects. We cannot. This is
9 9 ^(9){ }^{9}9 Recall that the square brackets here mean e β f e β f del_(e_(beta))f\partial_{\boldsymbol{e}_{\beta}} feβf or, in a coordinate basis f / x β f / x β del f//delx^(beta)\partial f / \partial x^{\beta}f/xβ and that this quantity is a number.
10 10 ^(10){ }^{10}10 We explain how to access the components S i 1 , i n j 1 j m ; k S i 1 , i n j 1 j m ; k S^(i_(1),dotsi_(n))_(j_(1)dotsj_(m);k)S^{i_{1}, \ldots i_{n}}{ }_{j_{1} \ldots j_{m} ; k}Si1,inj1jm;k in the next sections.

11 11 ^(11){ }^{11}11 By extension, we have

ω ν = Γ ν β μ ω ν ω β ω ν = Γ ν β μ ω ν ω β gradomega^(nu)=-Gamma^(nu)_(beta mu)omega^(nu)oxomega^(beta)\boldsymbol{\nabla} \boldsymbol{\omega}^{\nu}=-\Gamma^{\nu}{ }_{\beta \mu} \boldsymbol{\omega}^{\nu} \otimes \boldsymbol{\omega}^{\beta}ων=Γνβμωνωβ.
This should be contrasted with a result we shall establish in Chapter 36 that
(34.31) d ω ν = Γ ν β μ ω ν ω β . (34.31) d ω ν = Γ ν β μ ω ν ω β . {:(34.31)domega^(nu)=Gamma^(nu)_(beta mu)omega^(nu)^^omega^(beta).:}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\nu}=\Gamma^{\nu}{ }_{\beta \mu} \boldsymbol{\omega}^{\nu} \wedge \boldsymbol{\omega}^{\beta} . \tag{34.31} \end{equation*}(34.31)dων=Γνβμωνωβ.
12 12 ^(12){ }^{12}12 Notice how this expression has minus sign in front of the connection coefficients, in contrast to the vector version.
because the action of d d d\boldsymbol{d}d on a p p ppp-form produces a ( p + 1 ) ( p + 1 ) (p+1)(p+1)(p+1)-form and not the object whose components are the covariant derivatives. This can be traced back to the fact that the action of d d d\boldsymbol{d}d on p p ppp-forms with p 1 p 1 p >= 1p \geq 1p1 involves the antisymmetric wedge product ^^\wedge and not the tensor product ox\otimes. This presents a problem since the action of the connection operator on a tensor does not yield an antisymmetric object, but simply a tensor built from ordinary tensor products. In fact, for a general ( n , m ) ( n , m ) (n,m)(n, m)(n,m) tensor S S S\boldsymbol{S}S, we define
S = S i 1 , i n j 1 j m ; k ( e i 1 e i n ω j 1 ω j m ω k ) , S = S i 1 , i n j 1 j m ; k e i 1 e i n ω j 1 ω j m ω k , grad S=S^(i_(1),dotsi_(n))_(j_(1)dotsj_(m);k)(e_(i_(1))ox dots oxe_(i_(n))oxomega^(j_(1))ox dots oxomega^(j_(m))oxomega^(k)),quad\nabla \boldsymbol{S}=S^{i_{1}, \ldots i_{n}}{ }_{j_{1} \ldots j_{m} ; k}\left(\boldsymbol{e}_{i_{1}} \otimes \ldots \otimes \boldsymbol{e}_{i_{n}} \otimes \boldsymbol{\omega}^{j_{1}} \otimes \ldots \otimes \boldsymbol{\omega}^{j_{m}} \otimes \boldsymbol{\omega}^{k}\right), \quadS=Si1,inj1jm;k(ei1einωj1ωjmωk), (34.28)
which does not involve the wedge product. 10 10 ^(10){ }^{10}10 The point here is that we need a different approach to evaluate the covariant derivative of a p p ppp-form.
Example 34.7
Let's derive the action of the covariant derivative μ μ grad_(mu)\nabla_{\mu}μ on 1-forms. We use the same method we used in the last chapter. This involves combining the 1 -form with a vector via the inner product to make a number. The derivative of the combined vector and 1-form may be found using the Leibniz product rule for tensors.
This product rule for the connection is ( S T ) = S T + ( S T ) ( S T ) = S T + ( S T ) grad(S ox T)=grad S ox T+(S ox grad T)\boldsymbol{\nabla}(\boldsymbol{S} \otimes \boldsymbol{T})=\boldsymbol{\nabla} \boldsymbol{S} \otimes \boldsymbol{T}+(\boldsymbol{S} \otimes \boldsymbol{\nabla} \boldsymbol{T})(ST)=ST+(ST). Take S = ω ν S = ω ν S=omega^(nu)\boldsymbol{S}=\boldsymbol{\omega}^{\nu}S=ων and T = e μ T = e μ T=e_(mu)\boldsymbol{T}=\boldsymbol{e}_{\mu}T=eμ and replace the ox\otimes operation with (:\langle, . T h i s i s p e r m i s s i b l e a s . T h i s i s p e r m i s s i b l e a s :).Thisispermissibleas\rangle . This is permissible as.Thisispermissibleas the inner product preserves the linear structure enforced by tensor product ox\otimes. The derivative of ω ν , e μ = δ ν μ ω ν , e μ = δ ν μ (:omega^(nu),e_(mu):)=delta^(nu)_(mu)\left\langle\boldsymbol{\omega}^{\nu}, \boldsymbol{e}_{\mu}\right\rangle=\delta^{\nu}{ }_{\mu}ων,eμ=δνμ is zero, so we obtain
0 = ( ω ν , e μ ) = ω ν , e μ + ω ν , e μ = ω ν ( e μ , ) + Γ α β μ ω β ( ) e α ( ω ν ) (34.29) = ω ν ( e μ , ) + Γ α β μ ω β ( ) δ α ν , 0 = ω ν , e μ = ω ν , e μ + ω ν , e μ = ω ν e μ , + Γ α β μ ω β ( ) e α ω ν (34.29) = ω ν e μ , + Γ α β μ ω β ( ) δ α ν , {:[0=(:grad(omega^(nu),e_(mu)):)=(:gradomega^(nu),e_(mu):)+(:omega^(nu),grade_(mu):)],[=gradomega^(nu)(e_(mu),)+Gamma^(alpha)_(beta mu)omega^(beta)()oxe_(alpha)(omega^(nu))],[(34.29)=gradomega^(nu)(e_(mu),)+Gamma^(alpha)_(beta mu)omega^(beta)()delta_(alpha)^(nu)","]:}\begin{align*} 0=\left\langle\boldsymbol{\nabla}\left(\boldsymbol{\omega}^{\nu}, \boldsymbol{e}_{\mu}\right)\right\rangle & =\left\langle\boldsymbol{\nabla} \boldsymbol{\omega}^{\nu}, \boldsymbol{e}_{\mu}\right\rangle+\left\langle\boldsymbol{\omega}^{\nu}, \boldsymbol{\nabla} \boldsymbol{e}_{\mu}\right\rangle \\ & =\boldsymbol{\nabla} \boldsymbol{\omega}^{\nu}\left(\boldsymbol{e}_{\mu},\right)+\Gamma^{\alpha}{ }_{\beta \mu} \boldsymbol{\omega}^{\beta}() \otimes \boldsymbol{e}_{\alpha}\left(\boldsymbol{\omega}^{\nu}\right) \\ & =\boldsymbol{\nabla} \boldsymbol{\omega}^{\nu}\left(\boldsymbol{e}_{\mu},\right)+\Gamma^{\alpha}{ }_{\beta \mu} \boldsymbol{\omega}^{\beta}() \delta_{\alpha}^{\nu}, \tag{34.29} \end{align*}0=(ων,eμ)=ων,eμ+ων,eμ=ων(eμ,)+Γαβμωβ()eα(ων)(34.29)=ων(eμ,)+Γαβμωβ()δαν,
where, in the second line, we used e μ = Γ α β μ ω β e α e μ = Γ α β μ ω β e α grade_(mu)=Gamma^(alpha)_(beta mu)omega^(beta)oxe_(alpha)\nabla \boldsymbol{e}_{\mu}=\Gamma^{\alpha}{ }_{\beta \mu} \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{e}_{\alpha}eμ=Γαβμωβeα.
From the last example we conclude that 11 11 ^(11){ }^{11}11
(34.32) ω ν ( e μ , ) = Γ β μ ν ω β (34.32) ω ν e μ , = Γ β μ ν ω β {:(34.32)gradomega^(nu)(e_(mu),)=-Gamma_(beta mu)^(nu)omega^(beta):}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{\omega}^{\nu}\left(e_{\mu},\right)=-\Gamma_{\beta \mu}^{\nu} \omega^{\beta} \tag{34.32} \end{equation*}(34.32)ων(eμ,)=Γβμνωβ
We can use this to calculate the covariant derivative of the 1 -form
β σ ~ = ( β σ ν ) ω ν + σ ν β ω ν (34.33) = σ ν x β ω ν σ ν Γ β μ ν ω μ β σ ~ = β σ ν ω ν + σ ν β ω ν (34.33) = σ ν x β ω ν σ ν Γ β μ ν ω μ {:[grad_(beta) tilde(sigma)=(grad_(beta)sigma_(nu))omega^(nu)+sigma_(nu)grad_(beta)omega^(nu)],[(34.33)=(delsigma_(nu))/(delx^(beta))omega^(nu)-sigma_(nu)Gamma_(beta mu)^(nu)omega^(mu)]:}\begin{align*} \boldsymbol{\nabla}_{\beta} \tilde{\boldsymbol{\sigma}} & =\left(\boldsymbol{\nabla}_{\beta} \sigma_{\nu}\right) \boldsymbol{\omega}^{\nu}+\sigma_{\nu} \boldsymbol{\nabla}_{\beta} \boldsymbol{\omega}^{\nu} \\ & =\frac{\partial \sigma_{\nu}}{\partial x^{\beta}} \boldsymbol{\omega}^{\nu}-\sigma_{\nu} \Gamma_{\beta \mu}^{\nu} \boldsymbol{\omega}^{\mu} \tag{34.33} \end{align*}βσ~=(βσν)ων+σνβων(34.33)=σνxβωνσνΓβμνωμ
The result, on relabelling indices, is an equation for the action of the covariant derivative operator on a 1 -form 12 12 ^(12){ }^{12}12
(34.34) μ σ ~ = ( σ ν x μ σ λ Γ μ ν λ ) ω ν (34.34) μ σ ~ = σ ν x μ σ λ Γ μ ν λ ω ν {:(34.34)grad_(mu) tilde(sigma)=((delsigma_(nu))/(delx^(mu))-sigma_(lambda)Gamma_(mu nu)^(lambda))omega^(nu):}\begin{equation*} \nabla_{\mu} \tilde{\boldsymbol{\sigma}}=\left(\frac{\partial \sigma_{\nu}}{\partial x^{\mu}}-\sigma_{\lambda} \Gamma_{\mu \nu}^{\lambda}\right) \boldsymbol{\omega}^{\nu} \tag{34.34} \end{equation*}(34.34)μσ~=(σνxμσλΓμνλ)ων
We can also write an equation for the components in semicolon notation
(34.35) σ ν ; μ = σ ν , μ σ λ Γ μ ν λ (34.35) σ ν ; μ = σ ν , μ σ λ Γ μ ν λ {:(34.35)sigma_(nu;mu)=sigma_(nu,mu)-sigma_(lambda)Gamma_(mu nu)^(lambda):}\begin{equation*} \sigma_{\nu ; \mu}=\sigma_{\nu, \mu}-\sigma_{\lambda} \Gamma_{\mu \nu}^{\lambda} \tag{34.35} \end{equation*}(34.35)σν;μ=σν,μσλΓμνλ
or an equation for the 1-form that results if the direction of the derivative is given by a vector u u u\boldsymbol{u}u,
(34.36) ( u σ ~ ) α = u μ ( σ α x μ σ λ Γ μ α λ ) (34.36) u σ ~ α = u μ σ α x μ σ λ Γ μ α λ {:(34.36)(grad_(u)( tilde(sigma)))_(alpha)=u^(mu)((delsigma_(alpha))/(delx^(mu))-sigma_(lambda)Gamma_(mu alpha)^(lambda)):}\begin{equation*} \left(\nabla_{u} \tilde{\boldsymbol{\sigma}}\right)_{\alpha}=u^{\mu}\left(\frac{\partial \sigma_{\alpha}}{\partial x^{\mu}}-\sigma_{\lambda} \Gamma_{\mu \alpha}^{\lambda}\right) \tag{34.36} \end{equation*}(34.36)(uσ~)α=uμ(σαxμσλΓμαλ)
As shown in the next example, we can also express the result of applying the exterior derivative to 1 -forms in a coordinate frame in terms of the action of the covariant derivative.
Example 34.8
A 2 -form is obtained by acting on the 1-form A ~ = A μ d x μ A ~ = A μ d x μ tilde(A)=A_(mu)dx^(mu)\tilde{\boldsymbol{A}}=A_{\mu} \boldsymbol{d} x^{\mu}A~=Aμdxμ with the exterior derivative operator d d d\boldsymbol{d}d. This is written
(34.37) d A ~ = A μ x α d x α d x μ (34.37) d A ~ = A μ x α d x α d x μ {:(34.37)d tilde(A)=(delA_(mu))/(delx^(alpha))dx^(alpha)^^dx^(mu):}\begin{equation*} \boldsymbol{d} \tilde{\boldsymbol{A}}=\frac{\partial A_{\mu}}{\partial x^{\alpha}} \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\mu} \tag{34.37} \end{equation*}(34.37)dA~=Aμxαdxαdxμ
To point the derivative along the e β e β e_(beta)\boldsymbol{e}_{\beta}eβ direction, we contract, which is equivalent to filling the first slot of the 2 -form
d A ~ , e β = d A ~ ( e β , ) = A μ x α [ d x α d x μ ] ( e β , ) = A μ x α [ d x α d x μ ( e β , ) d x μ d x α ( e β , ) ] = A μ x α ( δ α β d x μ δ μ β d x α ) = A μ x β d x μ A β x α d x α (34.38) = ( A μ x β A β x μ ) d x μ d A ~ , e β = d A ~ e β , = A μ x α d x α d x μ e β , = A μ x α d x α d x μ e β , d x μ d x α e β , = A μ x α δ α β d x μ δ μ β d x α = A μ x β d x μ A β x α d x α (34.38) = A μ x β A β x μ d x μ {:[(:d( tilde(A)),e_(beta):)=d tilde(A)(e_(beta),)=(delA_(mu))/(delx^(alpha))[dx^(alpha)^^dx^(mu)](e_(beta),)],[=(delA_(mu))/(delx^(alpha))[dx^(alpha)ox dx^(mu)(e_(beta),)-dx^(mu)ox dx^(alpha)(e_(beta),)]],[=(delA_(mu))/(delx^(alpha))(delta^(alpha)_(beta)dx^(mu)-delta^(mu)_(beta)dx^(alpha))],[=(delA_(mu))/(delx^(beta))dx^(mu)-(delA_(beta))/(delx^(alpha))dx^(alpha)],[(34.38)=((delA_(mu))/(delx^(beta))-(delA_(beta))/(delx^(mu)))dx^(mu)]:}\begin{align*} \left\langle\boldsymbol{d} \tilde{\boldsymbol{A}}, \boldsymbol{e}_{\beta}\right\rangle=\boldsymbol{d} \tilde{\boldsymbol{A}}\left(\boldsymbol{e}_{\beta},\right) & =\frac{\partial A_{\mu}}{\partial x^{\alpha}}\left[\boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\mu}\right]\left(\boldsymbol{e}_{\beta},\right) \\ & =\frac{\partial A_{\mu}}{\partial x^{\alpha}}\left[\boldsymbol{d} x^{\alpha} \otimes \boldsymbol{d} x^{\mu}\left(\boldsymbol{e}_{\beta},\right)-\boldsymbol{d} x^{\mu} \otimes \boldsymbol{d} x^{\alpha}\left(\boldsymbol{e}_{\beta},\right)\right] \\ & =\frac{\partial A_{\mu}}{\partial x^{\alpha}}\left(\delta^{\alpha}{ }_{\beta} \boldsymbol{d} x^{\mu}-\delta^{\mu}{ }_{\beta} \boldsymbol{d} x^{\alpha}\right) \\ & =\frac{\partial A_{\mu}}{\partial x^{\beta}} \boldsymbol{d} x^{\mu}-\frac{\partial A_{\beta}}{\partial x^{\alpha}} \boldsymbol{d} x^{\alpha} \\ & =\left(\frac{\partial A_{\mu}}{\partial x^{\beta}}-\frac{\partial A_{\beta}}{\partial x^{\mu}}\right) \boldsymbol{d} x^{\mu} \tag{34.38} \end{align*}dA~,eβ=dA~(eβ,)=Aμxα[dxαdxμ](eβ,)=Aμxα[dxαdxμ(eβ,)dxμdxα(eβ,)]=Aμxα(δαβdxμδμβdxα)=AμxβdxμAβxαdxα(34.38)=(AμxβAβxμ)dxμ
This is a 1-form. Extract the γ γ gamma\gammaγ component by contracting with e γ e γ e_(gamma)\boldsymbol{e}_{\gamma}eγ to find
(34.39) d A ~ ( e β , e γ ) = ( A γ x β A β x γ ) (34.39) d A ~ e β , e γ = A γ x β A β x γ {:(34.39)d tilde(A)(e_(beta),e_(gamma))=((delA_(gamma))/(delx^(beta))-(delA_(beta))/(delx^(gamma))):}\begin{equation*} \boldsymbol{d} \tilde{\boldsymbol{A}}\left(\boldsymbol{e}_{\beta}, \boldsymbol{e}_{\gamma}\right)=\left(\frac{\partial A_{\gamma}}{\partial x^{\beta}}-\frac{\partial A_{\beta}}{\partial x^{\gamma}}\right) \tag{34.39} \end{equation*}(34.39)dA~(eβ,eγ)=(AγxβAβxγ)
Compare this to the coordinate-frame equation
A γ ; β A β ; γ = ( A γ x β Γ β γ σ A σ ) ( A β x γ Γ γ β σ A σ ) (34.40) = A γ x β A β x γ . A γ ; β A β ; γ = A γ x β Γ β γ σ A σ A β x γ Γ γ β σ A σ (34.40) = A γ x β A β x γ . {:[A_(gamma;beta)-A_(beta;gamma)=((delA_(gamma))/(delx^(beta))-Gamma_(beta gamma)^(sigma)A_(sigma))-((delA_(beta))/(delx^(gamma))-Gamma_(gamma beta)^(sigma)A_(sigma))],[(34.40)=(delA_(gamma))/(delx^(beta))-(delA_(beta))/(delx^(gamma)).]:}\begin{align*} A_{\gamma ; \beta}-A_{\beta ; \gamma} & =\left(\frac{\partial A_{\gamma}}{\partial x^{\beta}}-\Gamma_{\beta \gamma}^{\sigma} A_{\sigma}\right)-\left(\frac{\partial A_{\beta}}{\partial x^{\gamma}}-\Gamma_{\gamma \beta}^{\sigma} A_{\sigma}\right) \\ & =\frac{\partial A_{\gamma}}{\partial x^{\beta}}-\frac{\partial A_{\beta}}{\partial x^{\gamma}} . \tag{34.40} \end{align*}Aγ;βAβ;γ=(AγxβΓβγσAσ)(AβxγΓγβσAσ)(34.40)=AγxβAβxγ.
We see that we can write the components of the 2 -form d A ~ = 1 2 ( d A ~ ) β γ d x β d x γ d A ~ = 1 2 ( d A ~ ) β γ d x β d x γ d tilde(A)=(1)/(2)(d tilde(A))_(beta gamma)dx^(beta)^^dx^(gamma)\boldsymbol{d} \tilde{\boldsymbol{A}}=\frac{1}{2}(\boldsymbol{d} \tilde{\boldsymbol{A}})_{\beta \gamma} \boldsymbol{d} x^{\beta} \wedge \boldsymbol{d} x^{\gamma}dA~=12(dA~)βγdxβdxγ
as
( d A ~ ) β γ = ( A γ ; β A β ; γ ) = 2 A [ β ; γ ] , ( d A ~ ) β γ = A γ ; β A β ; γ = 2 A [ β ; γ ] , (d tilde(A))_(beta gamma)=(A_(gamma;beta)-A_(beta;gamma))=-2A_([beta;gamma]),(\boldsymbol{d} \tilde{\boldsymbol{A}})_{\beta \gamma}=\left(A_{\gamma ; \beta}-A_{\beta ; \gamma}\right)=-2 A_{[\beta ; \gamma]},(dA~)βγ=(Aγ;βAβ;γ)=2A[β;γ],
( 34.41 ) ( 34.41 ) (34.41)(34.41)(34.41)
where the square bracket notation for antisymmetrization has been used in the last line. 13 13 ^(13){ }^{13}13
We now extend the action of the covariant derivative from vectors and 1 -forms to tensors in general. Compare the vector v = v μ e μ v = v μ e μ v=v^(mu)e_(mu)\boldsymbol{v}=v^{\mu} \boldsymbol{e}_{\mu}v=vμeμ to the (2,0) tensor
(34.44) T = T μ ν e μ e ν (34.44) T = T μ ν e μ e ν {:(34.44)T=T^(mu nu)e_(mu)oxe_(nu):}\begin{equation*} \boldsymbol{T}=T^{\mu \nu} \boldsymbol{e}_{\mu} \otimes \boldsymbol{e}_{\nu} \tag{34.44} \end{equation*}(34.44)T=Tμνeμeν
The vector v v v\boldsymbol{v}v is made up of components v μ v μ v^(mu)v^{\mu}vμ and the basis vectors e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ and we saw that, when acted on by the derivative operator α α grad_(alpha)\nabla_{\alpha}α (i.e. the covariant derivative in the direction e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα ), we obtain a contribution from the components v μ / x α v μ / x α delv^(mu)//delx^(alpha)\partial v^{\mu} / \partial x^{\alpha}vμ/xα, added to a contribution from the basis vectors of Γ μ α β v β Γ μ α β v β Gamma^(mu)_(alpha beta)v^(beta)\Gamma^{\mu}{ }_{\alpha \beta} v^{\beta}Γμαβvβ. The tensor T T T\boldsymbol{T}T is very similar, only with an additional contribution from a second basis vector. We therefore obtain an additional connection Γ Γ Gamma\GammaΓ for this basis vector. The rule for the derivatives of a ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor are therefore, in component form,
(34.45) ( α T ) μ ν = T μ ν x α + Γ α β μ T β ν + Γ α β ν T μ β . (34.45) α T μ ν = T μ ν x α + Γ α β μ T β ν + Γ α β ν T μ β . {:(34.45)(grad_(alpha)T)^(mu nu)=(delT^(mu nu))/(delx^(alpha))+Gamma_(alpha beta)^(mu)T^(beta nu)+Gamma_(alpha beta)^(nu)T^(mu beta).:}\begin{equation*} \left(\boldsymbol{\nabla}_{\alpha} \boldsymbol{T}\right)^{\mu \nu}=\frac{\partial T^{\mu \nu}}{\partial x^{\alpha}}+\Gamma_{\alpha \beta}^{\mu} T^{\beta \nu}+\Gamma_{\alpha \beta}^{\nu} T^{\mu \beta} . \tag{34.45} \end{equation*}(34.45)(αT)μν=Tμνxα+ΓαβμTβν+ΓαβνTμβ.
13 13 ^(13){ }^{13}13 This expression is the basis of the useful equation for the 1 -form α ~ α ~ tilde(alpha)\tilde{\boldsymbol{\alpha}}α~ that says
d α ~ , u v = u α ~ , v v α ~ , u d α ~ , u v = u α ~ , v v α ~ , u (:d tilde(alpha),u^^v:)=(:grad_(u)( tilde(alpha)),v:)-(:grad_(v)( tilde(alpha)),u:)\langle\boldsymbol{d} \tilde{\boldsymbol{\alpha}}, \boldsymbol{u} \wedge \boldsymbol{v}\rangle=\left\langle\nabla_{u} \tilde{\boldsymbol{\alpha}}, \boldsymbol{v}\right\rangle-\left\langle\nabla_{v} \tilde{\boldsymbol{\alpha}}, \boldsymbol{u}\right\rangledα~,uv=uα~,vvα~,u.
(34.42)
This can be shown by writing the inner product in components:
( α | γ ; β | α | β ; γ | ) ( u β v γ u γ v β ) α | γ ; β | α | β ; γ | u β v γ u γ v β (alpha_(|gamma;beta|)-alpha_(|beta;gamma|))(u^(beta)v^(gamma)-u^(gamma)v^(beta))\left(\alpha_{|\gamma ; \beta|}-\alpha_{|\beta ; \gamma|}\right)\left(u^{\beta} v^{\gamma}-u^{\gamma} v^{\beta}\right)(α|γ;β|α|β;γ|)(uβvγuγvβ). (34.43)
The details are left as an exercise, and the expression is used in Chapter 43.
\curvearrowright Eqn 34.45 was originally claimed in eqn 12.37 in Chapter 12.
Fig. 34.2 A manifold with a metric provides enough structure to define an affine connection, and then curvature, and then the Einstein tensor. These are all required for general relativity.
14 14 ^(14){ }^{14}14 The Riemann tensor will be revisited in Chapter 35.
15 15 ^(15){ }^{15}15 This statement is the final word on what the spacetime of general relativity is, from the mathematical point of view. For those interested in the history, Jean le Rond d'Alembert (17171783) was probably the first person to 1783) was probably the first person to
express time as the fourth dimension and thereby invent the notion of spaceand thereby invent the notion of space-
time. This advance was attributed to time. This advance was attributed to
Lagrange by E. T. Bell, but there is litLagrange by E. T. Bell, but there is lit
tle evidence for this. See R. G. Van Oss, Historia Mathematica 10, 455 (1983) for a discussion.
Notice how the connection-coefficients are contracted against each tensor index in turn.
Since basis 1 -forms make a Γ a Γ a-Gamma\mathrm{a}-\GammaaΓ contribution, we similarly expect a contribution of one of these for each basis 1 -form. So, for the ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor ξ ξ xi\boldsymbol{\xi}ξ we have
(34.46) ( α ξ ) μ ν = ξ μ ν x α Γ α μ λ ξ λ ν Γ α ν λ ξ μ λ (34.46) α ξ μ ν = ξ μ ν x α Γ α μ λ ξ λ ν Γ α ν λ ξ μ λ {:(34.46)(grad_(alpha)xi)_(mu nu)=(delxi_(mu nu))/(delx^(alpha))-Gamma_(alpha mu)^(lambda)xi_(lambda nu)-Gamma_(alpha nu)^(lambda)xi_(mu lambda):}\begin{equation*} \left(\boldsymbol{\nabla}_{\alpha} \boldsymbol{\xi}\right)_{\mu \nu}=\frac{\partial \xi_{\mu \nu}}{\partial x^{\alpha}}-\Gamma_{\alpha \mu}^{\lambda} \xi_{\lambda \nu}-\Gamma_{\alpha \nu}^{\lambda} \xi_{\mu \lambda} \tag{34.46} \end{equation*}(34.46)(αξ)μν=ξμνxαΓαμλξλνΓανλξμλ
Again, we must contract the connection coefficients against each down index.
Generalizing, the routine for a general ( m , n ) ( m , n ) (m,n)(m, n)(m,n) tensor A A A\boldsymbol{A}A is simply to add additional Γ Γ Gamma\GammaΓ s for each up index and subtract them for each down index:
( α A ) β γ δ κ μ ν ξ ω = A β γ δ κ μ ν ξ ω x α + Γ α λ μ A β γ δ κ λ ν ξ ω + Γ ν α λ A β γ δ κ μ λ ξ ω + Γ ( terms for all up indices ) Γ λ α β A λ γ δ κ μ ν ξ ω Γ λ α γ A β λ δ κ μ ν ξ ω (34.47) Γ ( terms for all down indices ) . α A β γ δ κ μ ν ξ ω = A β γ δ κ μ ν ξ ω x α + Γ α λ μ A β γ δ κ λ ν ξ ω + Γ ν α λ A β γ δ κ μ λ ξ ω + Γ (  terms for all up indices  ) Γ λ α β A λ γ δ κ μ ν ξ ω Γ λ α γ A β λ δ κ μ ν ξ ω (34.47) Γ (  terms for all down indices  ) . {:[(grad_(alpha)A)_(beta gamma delta dots kappa)^(mu nu xi dots omega)=(delA_(beta gamma delta dots kappa)^(mu nu xi dots omega))/(delx^(alpha))+Gamma_(alpha lambda)^(mu)A_(beta gamma delta dots kappa)^(lambda nu xi dots omega)+Gamma^(nu)_(alpha lambda)A_(beta gamma delta dots kappa)^(mu lambda xi dots omega)],[+Gamma(" terms for all up indices ")],[-Gamma^(lambda)_(alpha beta)A_(lambda gamma delta dots kappa)^(mu nu xi dots omega)-Gamma^(lambda)_(alpha gamma)A_(beta lambda delta dots kappa)^(mu nu xi dots omega)],[(34.47)-Gamma(" terms for all down indices ").]:}\begin{align*} \left(\nabla_{\alpha} A\right)_{\beta \gamma \delta \ldots \kappa}^{\mu \nu \xi \ldots \omega}= & \frac{\partial A_{\beta \gamma \delta \ldots \kappa}^{\mu \nu \xi \ldots \omega}}{\partial x^{\alpha}}+\Gamma_{\alpha \lambda}^{\mu} A_{\beta \gamma \delta \ldots \kappa}^{\lambda \nu \xi \ldots \omega}+\Gamma^{\nu}{ }_{\alpha \lambda} A_{\beta \gamma \delta \ldots \kappa}^{\mu \lambda \xi \ldots \omega} \\ & +\Gamma(\text { terms for all up indices }) \\ & -\Gamma^{\lambda}{ }_{\alpha \beta} A_{\lambda \gamma \delta \ldots \kappa}^{\mu \nu \xi \ldots \omega}-\Gamma^{\lambda}{ }_{\alpha \gamma} A_{\beta \lambda \delta \ldots \kappa}^{\mu \nu \xi \ldots \omega} \\ & -\Gamma(\text { terms for all down indices }) . \tag{34.47} \end{align*}(αA)βγδκμνξω=Aβγδκμνξωxα+ΓαλμAβγδκλνξω+ΓναλAβγδκμλξω+Γ( terms for all up indices )ΓλαβAλγδκμνξωΓλαγAβλδκμνξω(34.47)Γ( terms for all down indices ).
This completes the description of the connection and covariant derivative. Of course, the reason we need all of this formalism is that v v grad_(v)\nabla_{v}v is the derivative that is most useful in describing the curvature of spacetime encoded into the metric field. We make this link between the covariant derivative to the metric in the next and final section of this chapter.

34.4 The metric revisited

The central pillar of our geometric theory of Nature is the metric, the fundamental field of the theory of general relativity. Both the connection grad\boldsymbol{\nabla} and the Riemann tensor 14 R 14 R ^(14)R{ }^{14} \boldsymbol{R}14R can be thought of as structures that derive from the metric g g g\boldsymbol{g}g defined on the manifold M M M\mathcal{M}M, and these structures underpin general relativity (this idea is described schematically in Fig. 34.2). To specify these essential ingredients we write ( M , g ) ( M , g ) (M,g)(\mathcal{M}, \boldsymbol{g})(M,g). A mathematical statement of general relativity is then as follows: 15 15 ^(15){ }^{15}15
Spacetime is the manifold M M M\mathcal{M}M on which there is a Lorentz metric g g g\boldsymbol{g}g. The curvature of spacetime, described by g g g\boldsymbol{g}g, is related to the distribution of matter in spacetime by the Einstein equation.
As we've seen previously in this book, the metric tensor is defined mathematically to be a non-singular ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) tensor with the property g ( e μ , e ν ) = g μ ν g e μ , e ν = g μ ν g(e_(mu),e_(nu))=g_(mu nu)\boldsymbol{g}\left(\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right)=g_{\mu \nu}g(eμ,eν)=gμν. The definition implies that the metric tensor can be built from the tensor product of basis 1-forms: g = g μ ν ω μ ω ν g = g μ ν ω μ ω ν g=g_(mu nu)omega^(mu)oxomega^(nu)\boldsymbol{g}=g_{\mu \nu} \boldsymbol{\omega}^{\mu} \otimes \boldsymbol{\omega}^{\nu}g=gμνωμων, or, introducing some alternative notation, we can write things in terms of a line element tensor, d s 2 = g μ ν d x μ d x ν d s 2 = g μ ν d x μ d x ν ds^(2)=g_(mu nu)dx^(mu)ox dx^(nu)\boldsymbol{d} s^{2}=g_{\mu \nu} \boldsymbol{d} x^{\mu} \otimes \boldsymbol{d} x^{\nu}ds2=gμνdxμdxν. Written in slot machine form, we have
(34.48) d s 2 ( , ) g ( , ) = g μ ν d x μ ( ) d x ν ( ) . (34.48) d s 2 ( , ) g ( , ) = g μ ν d x μ ( ) d x ν ( ) . {:(34.48)ds^(2)(",")-=g(",")=g_(mu nu)dx^(mu)()ox dx^(nu)().:}\begin{equation*} \boldsymbol{d} s^{2}(,) \equiv \boldsymbol{g}(,)=g_{\mu \nu} \boldsymbol{d} x^{\mu}() \otimes \boldsymbol{d} x^{\nu}() . \tag{34.48} \end{equation*}(34.48)ds2(,)g(,)=gμνdxμ()dxν().
With the metric in our toolkit, we can show how possessing a connection grad\boldsymbol{\nabla} allows us access to the notion of parallelism and parallel transport. One possible version of parallel transport might be that the tangent vector u u u\boldsymbol{u}u to a geodesic, parametrized by λ λ lambda\lambdaλ, is moved along the curve and simply returns a vector proportional to that tangent vector 16 16 ^(16){ }^{16}16
(34.49) u u D u d λ u (34.49) u u D u d λ u {:(34.49)grad_(u)u-=(Du)/((d)lambda)prop u:}\begin{equation*} \nabla_{u} \boldsymbol{u} \equiv \frac{\mathrm{D} \boldsymbol{u}}{\mathrm{~d} \lambda} \propto \boldsymbol{u} \tag{34.49} \end{equation*}(34.49)uuDu dλu
A connection that obeys this, very general, statement of parallelism is called a non-affine connection. Such connections are characterized by parametrizations where λ λ lambda\lambdaλ does not mark of regular intervals along the curve, in Fig 34.3 (bottom). When a connection is non-affine the vector u u u\boldsymbol{u}u can get longer or shorter as it moves around, but always remains parallel to itself.
In contrast, an affine connection, where λ λ lambda\lambdaλ marks off regular intervals along the path, is characterized by the much more restrictive statement of parallelism that, for a geodesic parametrized by λ λ lambda\lambdaλ, we have 17 17 ^(17){ }^{17}17
(34.50) u u = D u d λ = 0 (34.50) u u = D u d λ = 0 {:(34.50)grad_(u)u=(Du)/((d)lambda)=0:}\begin{equation*} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}=\frac{\mathrm{D} \boldsymbol{u}}{\mathrm{~d} \lambda}=0 \tag{34.50} \end{equation*}(34.50)uu=Du dλ=0
which we recognize as the geodesic equation. When this condition holds we have the state of affairs we have previously described as parallel transport. The tangent vector is transported along the geodesic remaining parallel to itself and, crucially, its length remains constant (Fig 34.3, top). Tangent vectors of constant length are something we certainly want from our physical theory, since we require, for example, that the velocity vector u u u\boldsymbol{u}u, which is tangent to the world line of a massive particle, has a constant magnitude such that u u = 1 u u = 1 u*u=-1\boldsymbol{u} \cdot \boldsymbol{u}=-1uu=1.
It is the metric that guarantees that the connection is affine. As a consequence, we must now permanently fix the connection and the metric together. This link between connection and metric is made with the compatibility condition 18 18 ^(18){ }^{18}18
(34.51) g = 0 (34.51) g = 0 {:(34.51)grad g=0:}\begin{equation*} \nabla \boldsymbol{g}=0 \tag{34.51} \end{equation*}(34.51)g=0
This equation implies that the covariant derivative u g = 0 u g = 0 grad_(u)g=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{g}=0ug=0 when taken along a curve with tangent vector u u u\boldsymbol{u}u. As we show in the example below, this condition is enough to ensure that the length of any vector is a constant when it is parallel transported.
Example 34.9
The vectors v v v\boldsymbol{v}v and w w w\boldsymbol{w}w are parallel transported along the curve whose tangent is u u u\boldsymbol{u}u. That is
(34.52) u v = 0 , u w = 0 , u α v β ; α = 0 , u α w γ ; α = 0 . (34.52) u v = 0 , u w = 0 , u α v β ; α = 0 , u α w γ ; α = 0 . {:[(34.52)grad_(u)v=0","quadgrad_(u)w=0","],[u^(alpha)v^(beta)_(;alpha)=0","quadu^(alpha)w^(gamma)_(;alpha)=0.]:}\begin{gather*} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{v}=0, \quad \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{w}=0, \tag{34.52}\\ u^{\alpha} v^{\beta}{ }_{; \alpha}=0, \quad u^{\alpha} w^{\gamma}{ }_{; \alpha}=0 . \end{gather*}(34.52)uv=0,uw=0,uαvβ;α=0,uαwγ;α=0.
If the lengths of the vectors are unchanged by this operation then the inner product g ( v , w ) g ( v , w ) g(v,w)\boldsymbol{g}(\boldsymbol{v}, \boldsymbol{w})g(v,w) cannot change on parallel transportation. As a result we have
(34.53) u g ( v , w ) = 0 (34.53) u g ( v , w ) = 0 {:(34.53)grad_(u)g(v","w)=0:}\begin{equation*} \boldsymbol{\nabla}_{u} \boldsymbol{g}(\boldsymbol{v}, \boldsymbol{w})=0 \tag{34.53} \end{equation*}(34.53)ug(v,w)=0
16 16 ^(16){ }^{16}16 As in Part II, we again use the D / d λ D / d λ D//dlambda\mathrm{D} / \mathrm{d} \lambdaD/dλ notation here to denote a covariant derivative taken along the curve parametrized by λ λ lambda\lambdaλ in a spacetime with a connection.
17 17 ^(17){ }^{17}17 In Exercise 34.7, we show that if a parametrization λ λ lambda\lambdaλ that obeys this equation, then an alternative parametrization s = a λ + b s = a λ + b s=a lambda+bs=a \lambda+bs=aλ+b, with a a aaa and b b bbb constants, also obeys the equation.
Fig. 34.3 Affine and non-affine connections.
18 18 ^(18){ }^{18}18 We note that there is an approach to gravity where the connection and the metric are treated as independent variables, with respect to which the Einstein-Hilbert action is varied (see Chapter 40). This is known as Palatini gravity. However, if the action is given by eqn 40.36 , formed from the (not necessarily metric-compatible) connection, then the two versions of gravity are equivalent.
or, written in components
Using the Leibniz rule: u α ( g β γ v β w γ ) ; α = 0 u α g β γ v β w γ ; α = 0 quadu^(alpha)(g_(beta gamma)v^(beta)w^(gamma))_(;alpha)=0\quad u^{\alpha}\left(g_{\beta \gamma} v^{\beta} w^{\gamma}\right)_{; \alpha}=0uα(gβγvβwγ);α=0.
u α g β γ ; α v β w γ + u α g β γ v β ; α w γ + u α g β γ v β w γ ; α = 0 u α g β γ ; α v β w γ + u α g β γ v β ; α w γ + u α g β γ v β w γ ; α = 0 quadu^(alpha)g_(beta gamma;alpha)v^(beta)w^(gamma)+u^(alpha)g_(beta gamma)v^(beta)_(;alpha)w^(gamma)+u^(alpha)g_(beta gamma)v^(beta)w^(gamma)_(;alpha)=0\quad u^{\alpha} g_{\beta \gamma ; \alpha} v^{\beta} w^{\gamma}+u^{\alpha} g_{\beta \gamma} v^{\beta}{ }_{; \alpha} w^{\gamma}+u^{\alpha} g_{\beta \gamma} v^{\beta} w^{\gamma}{ }_{; \alpha}=0uαgβγ;αvβwγ+uαgβγvβ;αwγ+uαgβγvβwγ;α=0.
(34.55) u α g β γ ; α v β w γ + u α g β γ v ; α β w γ + u α g β γ v β w γ ; α = 0 (34.55) u α g β γ ; α v β w γ + u α g β γ v ; α β w γ + u α g β γ v β w γ ; α = 0 {:(34.55)u^(alpha)g_(beta gamma;alpha)v^(beta)w^(gamma)+u^(alpha)g_(beta gamma)v_(;alpha)^(beta)w^(gamma)+u^(alpha)g_(beta gamma)v^(beta)w^(gamma)_(;alpha)=0:}\begin{equation*} u^{\alpha} g_{\beta \gamma ; \alpha} v^{\beta} w^{\gamma}+u^{\alpha} g_{\beta \gamma} v_{; \alpha}^{\beta} w^{\gamma}+u^{\alpha} g_{\beta \gamma} v^{\beta} w^{\gamma}{ }_{; \alpha}=0 \tag{34.55} \end{equation*}(34.55)uαgβγ;αvβwγ+uαgβγv;αβwγ+uαgβγvβwγ;α=0
The parallel transport conditions in eqn 34.52 kill the latter two terms on the left, and we obtain
This holds if, and only if,
t α v β w γ g β γ ; α = 0 t α v β w γ g β γ ; α = 0 t^(alpha)v^(beta)w^(gamma)g_(beta gamma;alpha)=0t^{\alpha} v^{\beta} w^{\gamma} g_{\beta \gamma ; \alpha}=0tαvβwγgβγ;α=0
g β γ ; α = 0 . g β γ ; α = 0 . g_(beta gamma;alpha)=0.g_{\beta \gamma ; \alpha}=0 .gβγ;α=0.
The components of g g grad g\boldsymbol{\nabla} \boldsymbol{g}g are all zero therefore, and so this object vanishes.
As a special case, we also see that for a geodesic with tangent u u u\boldsymbol{u}u, then fixing g β γ ; α = 0 g β γ ; α = 0 g_(beta gamma;alpha)=0g_{\beta \gamma ; \alpha}=0gβγ;α=0 and g ( u , u ) = 1 g ( u , u ) = 1 g(u,u)=-1\boldsymbol{g}(\boldsymbol{u}, \boldsymbol{u})=-1g(u,u)=1, ensures the geodesic equation u u u ν u μ ; ν = 0 u u u ν u μ ; ν = 0 grad_(u)u-=u^(nu)u^(mu)_(;nu)=0\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u} \equiv u^{\nu} u^{\mu}{ }_{; \nu}=0uuuνuμ;ν=0 is obeyed, as required. 19 19 ^(19){ }^{19}19
This makes the crucial link between the connection and the metric. The condition g = 0 g = 0 grad g=0\boldsymbol{\nabla} \boldsymbol{g}=0g=0 (along with the vanishing of torsion) is enough the fix the action of grad\nabla as the operator giving an affine connection. The reason that the covariant derivative is the suitable derivative for relativity is because it is the derivative that is built on the foundation of parallelism, which involved having access to a metric, and general relativity is the field theory of the metric field.
On a more practical note, metric components are useful in raising and lowering indices. The semicolon notation provides the means for doing this in expressions involving the covariant derivative, as we examine below.

Example 34.10

We act on the components of the covariant derivative with the metric components. We have
g μ ν v ; α μ = g μ ν ( v μ , α + Γ μ α λ v λ ) = v ν , α + Γ ν α λ v λ (34.58) = v ν , α + g λ γ Γ ν α λ v γ . g μ ν v ; α μ = g μ ν v μ , α + Γ μ α λ v λ = v ν , α + Γ ν α λ v λ (34.58) = v ν , α + g λ γ Γ ν α λ v γ . {:[g_(mu nu)v_(;alpha)^(mu)=g_(mu nu)(v^(mu)_(,alpha)+Gamma^(mu)_(alpha lambda)v^(lambda))],[=v_(nu,alpha)+Gamma_(nu alpha lambda)v^(lambda)],[(34.58)=v_(nu,alpha)+g^(lambda gamma)Gamma_(nu alpha lambda)v_(gamma).]:}\begin{align*} g_{\mu \nu} v_{; \alpha}^{\mu} & =g_{\mu \nu}\left(v^{\mu}{ }_{, \alpha}+\Gamma^{\mu}{ }_{\alpha \lambda} v^{\lambda}\right) \\ & =v_{\nu, \alpha}+\Gamma_{\nu \alpha \lambda} v^{\lambda} \\ & =v_{\nu, \alpha}+g^{\lambda \gamma} \Gamma_{\nu \alpha \lambda} v_{\gamma} . \tag{34.58} \end{align*}gμνv;αμ=gμν(vμ,α+Γμαλvλ)=vν,α+Γναλvλ(34.58)=vν,α+gλγΓναλvγ.
This last equation causes us to pause since Γ Γ Gamma\GammaΓ is not a tensor in its latter two down components, and so we cannot simply raise an index as we might be tempted to. This isn't a problem if we simply follow eqn 34.35 and write
g μ ν v μ ; α = v ν ; α = v ν , α Γ γ α ν v γ g μ ν v μ ; α = v ν ; α = v ν , α Γ γ α ν v γ g_(mu nu)v^(mu)_(;alpha)=v_(nu;alpha)=v_(nu,alpha)-Gamma^(gamma)_(alpha nu)v_(gamma)g_{\mu \nu} v^{\mu}{ }_{; \alpha}=v_{\nu ; \alpha}=v_{\nu, \alpha}-\Gamma^{\gamma}{ }_{\alpha \nu} v_{\gamma}gμνvμ;α=vν;α=vν,αΓγανvγ,
which gives us a way to interpret the final line in eqn 34.58 .

Chapter summary

  • The connection grad\boldsymbol{\nabla} allows a derivative to be defined that is satisfactory to describe tensor fields in the curved spacetimes of general relativity.
  • The components of grad\nabla are the connection coefficients Γ μ α β Γ μ α β Gamma^(mu)_(alpha beta)\Gamma^{\mu}{ }_{\alpha \beta}Γμαβ.
  • An affine connection is guaranteed by the condition g = 0 g = 0 grad g=0\nabla \boldsymbol{g}=0g=0.

Exercises

(34.1) Use the symmetry of the covariant derivative to show
(34.60) u + n v = u v + n v (34.60) u + n v = u v + n v {:(34.60)grad_(u+n)v=grad_(u)v+grad_(n)v:}\begin{equation*} \nabla_{u+n} v=\nabla_{u} v+\nabla_{n} v \tag{34.60} \end{equation*}(34.60)u+nv=uv+nv
(34.2) (a) Using the definition e α e β = g α β e α e β = g α β e_(alpha)*e_(beta)=g_(alpha beta)\boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta}=g_{\alpha \beta}eαeβ=gαβ, along with the action of grad\boldsymbol{\nabla} on basis vectors, to compute the derivative ( e α e β ) e α e β grad(e_(alpha)*e_(beta))\boldsymbol{\nabla}\left(\boldsymbol{e}_{\alpha} \cdot \boldsymbol{e}_{\beta}\right)(eαeβ), and use this to show
(34.61) g α β , σ = Γ σ α μ g μ β + Γ σ β μ g μ α . (34.61) g α β , σ = Γ σ α μ g μ β + Γ σ β μ g μ α . {:(34.61)g_(alpha beta,sigma)=Gamma_(sigma alpha)^(mu)g_(mu beta)+Gamma_(sigma beta)^(mu)g_(mu alpha).:}\begin{equation*} g_{\alpha \beta, \sigma}=\Gamma_{\sigma \alpha}^{\mu} g_{\mu \beta}+\Gamma_{\sigma \beta}^{\mu} g_{\mu \alpha} . \tag{34.61} \end{equation*}(34.61)gαβ,σ=Γσαμgμβ+Γσβμgμα.
(b) Use the result of (a) to prove the expression
(34.62) Γ α β σ = 1 2 ( g α β , σ + g α σ , β g β σ , α ) , (34.62) Γ α β σ = 1 2 g α β , σ + g α σ , β g β σ , α , {:(34.62)Gamma_(alpha beta sigma)=(1)/(2)(g_(alpha beta,sigma)+g_(alpha sigma,beta)-g_(beta sigma,alpha))",":}\begin{equation*} \Gamma_{\alpha \beta \sigma}=\frac{1}{2}\left(g_{\alpha \beta, \sigma}+g_{\alpha \sigma, \beta}-g_{\beta \sigma, \alpha}\right), \tag{34.62} \end{equation*}(34.62)Γαβσ=12(gαβ,σ+gασ,βgβσ,α),
that links the connection coefficients to the derivatives of the components of g g g\boldsymbol{g}g.
(34.3) (a) For an arbitrary matrix M M _ M_\underline{\boldsymbol{M}}M there is an identity
(34.63) x λ ln det M ( x ) = Tr { M 1 ( x ) x λ M ( x ) } (34.63) x λ ln det M _ ( x ) = Tr M _ 1 ( x ) x λ M _ ( x ) {:(34.63)(del)/(delx^(lambda))*ln detM_(x)=Tr{M_^(-1)(x)(del)/(delx^(lambda))M_(x)}:}\begin{equation*} \frac{\partial}{\partial x^{\lambda}} \cdot \ln \operatorname{det} \underline{\boldsymbol{M}}(x)=\operatorname{Tr}\left\{\underline{\boldsymbol{M}}^{-1}(x) \frac{\partial}{\partial x^{\lambda}} \underline{\boldsymbol{M}}(x)\right\} \tag{34.63} \end{equation*}(34.63)xλlndetM(x)=Tr{M1(x)xλM(x)}
Prove this by considering a variation in ln ( det M ) ln ( det M _ ) ln(detM_)\ln (\operatorname{det} \underline{\boldsymbol{M}})ln(detM) owing to a variation δ x λ δ x λ deltax^(lambda)\delta x^{\lambda}δxλ that gives
δ ln det M = ln det ( M + δ M ) ln det M . δ ln det M _ = ln det ( M _ + δ M _ ) ln det M _ . delta ln detM_=ln det(M_+deltaM_)-ln detM_.\delta \ln \operatorname{det} \underline{\boldsymbol{M}}=\ln \operatorname{det}(\underline{\boldsymbol{M}}+\delta \underline{\boldsymbol{M}})-\ln \operatorname{det} \underline{\boldsymbol{M}} .δlndetM=lndet(M+δM)lndetM.
(b) Use the identity
(34.65) Γ λ μ μ = 1 2 g μ ρ g ρ μ x λ (34.65) Γ λ μ μ = 1 2 g μ ρ g ρ μ x λ {:(34.65)Gamma_(lambda mu)^(mu)=(1)/(2)g^(mu rho)(delg_(rho mu))/(delx^(lambda)):}\begin{equation*} \Gamma_{\lambda \mu}^{\mu}=\frac{1}{2} g^{\mu \rho} \frac{\partial g_{\rho \mu}}{\partial x^{\lambda}} \tag{34.65} \end{equation*}(34.65)Γλμμ=12gμρgρμxλ
along with the expression in part (a) to show
(34.66) Γ λ μ μ = 1 | g | x λ | g | (34.66) Γ λ μ μ = 1 | g | x λ | g | {:(34.66)Gamma_(lambda mu)^(mu)=(1)/(sqrt(|g|))(del)/(delx^(lambda))sqrt(|g|):}\begin{equation*} \Gamma_{\lambda \mu}^{\mu}=\frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x^{\lambda}} \sqrt{|g|} \tag{34.66} \end{equation*}(34.66)Γλμμ=1|g|xλ|g|
(c) Use this to show further that the divergence is given by
(34.67) v ; μ μ = 1 | g | x μ ( | g | v μ ) . (34.67) v ; μ μ = 1 | g | x μ | g | v μ . {:(34.67)v_(;mu)^(mu)=(1)/(sqrt(|g|))(del)/(delx^(mu))*(sqrt(|g|)v^(mu)).:}\begin{equation*} v_{; \mu}^{\mu}=\frac{1}{\sqrt{|g|}} \frac{\partial}{\partial x^{\mu}} \cdot\left(\sqrt{|g|} v^{\mu}\right) . \tag{34.67} \end{equation*}(34.67)v;μμ=1|g|xμ(|g|vμ).
(34.4) Using components, take the covariant derivative u u grad_(u)\boldsymbol{\nabla}_{u}u of an inner product of vector and 1-form σ ~ , v = f σ ~ , v = f (: tilde(sigma),v:)=f\langle\tilde{\boldsymbol{\sigma}}, \boldsymbol{v}\rangle=fσ~,v=f, where f f fff is a function, and use this to prove the rule for the components of the covariant derivative of a 1 -form.
(34.5) (a) If λ λ lambda\lambdaλ is an affine parameter such that u α = d x α d u α = d x α d u^(alpha)=(dx^(alpha))/(d)u^{\alpha}=\frac{\mathrm{d} x^{\alpha}}{\mathrm{d}}uα=dxαd, show that
(34.68) ( u u ) e 1 = d u 1 d λ u α u σ Γ α 1 σ . (34.68) u u e 1 = d u 1 d λ u α u σ Γ α 1 σ . {:(34.68)(grad_(u)u)*e_(1)=(du_(1))/((d)lambda)-u^(alpha)u_(sigma)Gamma_(alpha1)^(sigma).:}\begin{equation*} \left(\boldsymbol{\nabla}_{u} \boldsymbol{u}\right) \cdot \boldsymbol{e}_{1}=\frac{\mathrm{d} u_{1}}{\mathrm{~d} \lambda}-u^{\alpha} u_{\sigma} \Gamma_{\alpha 1}^{\sigma} . \tag{34.68} \end{equation*}(34.68)(uu)e1=du1 dλuαuσΓα1σ.
(b) Using this, show that if the metric functions g μ ν g μ ν g_(mu nu)g_{\mu \nu}gμν are independent of a coordinate x 1 x 1 x^(1)x^{1}x1 then u 1 u 1 u_(1)u_{1}u1 is a constant along the particle's geodesic world line. This amounts to the rule for finding Killing vectors that we discussed in the last chapter.
(34.6) The exterior derivative of a 1 -form
Consider the number formed by filling the slots of a 2-form as follows: d A ~ ( u , v ) d A ~ ( u , v ) d tilde(A)(u,v)\boldsymbol{d} \tilde{\boldsymbol{A}}(\boldsymbol{u}, \boldsymbol{v})dA~(u,v), where A ~ A ~ tilde(A)\tilde{\boldsymbol{A}}A~ is a 1-form and u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are vectors.
(a) Show that when there is a connection, we can express this quantity as
(34.69) d A ~ ( u , v ) = u A ~ ( v ) v A ~ ( u ) (34.69) d A ~ ( u , v ) = u A ~ ( v ) v A ~ ( u ) {:(34.69)d tilde(A)(u","v)=grad_(u) tilde(A)(v)-grad_(v) tilde(A)(u):}\begin{equation*} d \tilde{A}(u, v)=\nabla_{u} \tilde{A}(v)-\nabla_{v} \tilde{A}(u) \tag{34.69} \end{equation*}(34.69)dA~(u,v)=uA~(v)vA~(u)
(b) Show that this can also be expressed as
d A ~ ( u , v ) = u A ~ , v v A ~ , u A ~ ( [ u , v ] ) d A ~ ( u , v ) = u A ~ , v v A ~ , u A ~ ( [ u , v ] ) d tilde(A)(u,v)=grad_(u)(: tilde(A),v:)-grad_(v)(: tilde(A),u:)- tilde(A)([u,v])d \tilde{A}(u, v)=\nabla_{u}\langle\tilde{A}, v\rangle-\nabla_{v}\langle\tilde{A}, u\rangle-\tilde{A}([u, v])dA~(u,v)=uA~,vvA~,uA~([u,v])
( 34.70 ) ( 34.70 ) (34.70)(34.70)(34.70)
This suggests another route to eqn 34.69. Since we know that the action of the exterior and the covariant derivative on a scalar are identical, we could have started by considering the candidate quantity u A ~ , v u A ~ , v grad_(u)(: tilde(A),v:)\nabla_{\boldsymbol{u}}\langle\tilde{\boldsymbol{A}}, \boldsymbol{v}\rangleuA~,v and then antisymmetrized it. We then expand the resulting candidate equation for d A ( u , v ) d A ( u , v ) dA(u,v)\boldsymbol{d} \boldsymbol{A}(\boldsymbol{u}, \boldsymbol{v})dA(u,v) of u A , v v A , u u A , v v A , u grad_(u)(:A,v:)-grad_(v)(:A,u:)\boldsymbol{\nabla}_{\boldsymbol{u}}\langle\boldsymbol{A}, \boldsymbol{v}\rangle-\boldsymbol{\nabla}_{\boldsymbol{v}}\langle\boldsymbol{A}, \boldsymbol{u}\rangleuA,vvA,u and obtain
(34.71) u A ~ ( v ) v A ~ ( u ) + A ~ ( [ u , v ] ) (34.71) u A ~ ( v ) v A ~ ( u ) + A ~ ( [ u , v ] ) {:(34.71)grad_(u) tilde(A)(v)-grad_(v) tilde(A)(u)+ tilde(A)([u","v]):}\begin{equation*} \nabla_{u} \tilde{A}(v)-\nabla_{v} \tilde{A}(u)+\tilde{A}([u, v]) \tag{34.71} \end{equation*}(34.71)uA~(v)vA~(u)+A~([u,v])
Since there should be no dependence on the choice of vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v, we would then need to subtract the part of the resulting expression that depends on [ u , v ] [ u , v ] [u,v][\boldsymbol{u}, \boldsymbol{v}][u,v], since this encodes the Lie derivative of the vectors.
(c) Starting from the expression from part (b) that we have now 'derived', show that
(34.72) d A ~ = μ A ν ( d x μ d x ν ) (34.72) d A ~ = μ A ν d x μ d x ν {:(34.72)d tilde(A)=del_(mu)A_(nu)(dx^(mu)^^dx^(nu)):}\begin{equation*} \boldsymbol{d} \tilde{\boldsymbol{A}}=\partial_{\mu} A_{\nu}\left(\boldsymbol{d} x^{\mu} \wedge \boldsymbol{d} x^{\nu}\right) \tag{34.72} \end{equation*}(34.72)dA~=μAν(dxμdxν)
Hint: Assume the arbitrary vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v in (b) are constant in space.
(34.7) Consider a path parametrized by an affine parameter λ λ lambda\lambdaλ. Define a new parametrization s = f ( λ ) s = f ( λ ) s=f(lambda)s=f(\lambda)s=f(λ). Show that this makes no difference to the geodesic equation if s s sss and λ λ lambda\lambdaλ are linearly related.
(34.8) Consider a non-geodesic curve in a spacetime with an affine connection. Show that the compatibility condition implies that if the magnitude of the tangent vector u u u\boldsymbol{u}u is a constant along the curve, then u ( u u ) = 0 u u u = 0 u*(grad_(u)u)=0\boldsymbol{u} \cdot\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{u}\right)=0u(uu)=0.
Physically, this means that an affine parametriza-
tion that keeps the magnitude of the velocity vector constant for any curve, has a u u u\boldsymbol{u}u that is perpendicular to the acceleration a = ( u u ) a = u u a=(grad_(u)u)\boldsymbol{a}=\left(\boldsymbol{\nabla}_{u} \boldsymbol{u}\right)a=(uu), just as we have in the flat spacetime of special relativity. See Needham (2021) for a discussion of the material in this problem.

Riemann curvature revisited

Abstract

After Riemann had made known his discoveries, mathematicians busied themselves with working out his system of geometrical ideas formally; chief among these were Christoffel, Ricci, and Levi-Civita. Riemann ... clearly left the real development of his ideas in the hands of some subsequent scientist whose genius as a physicist could rise to equal flights with his own as a mathematician. After a lapse of seventy years this mission has been fulfilled by Einstein. Hermann Weyl (1885-1955) Space - Time - Matter

You wake up in a space station and feel a force keeping you in bed. If you can't tell whether this is because there's a gravitational field present or because the station is accelerating, then there's a simple experiment to carry out. Allow two particles to fall freely, starting them in motion on parallel paths, and later measure their relative acceleration. If this acceleration is non-zero it suggests the presence of a genuine curvature in spacetime. This experiment, a measurement of geodesic deviation, is a true test of spacetime curvature and hence of real gravitational fields, as it allows access to the Riemann tensor R R R\boldsymbol{R}R. It is this tensor that provides the key to assessing whether spacetime is curved.
In this chapter, we revisit the Riemann tensor and investigate the geometrical method of calculating R R R\boldsymbol{R}R. The tools we need are the Lie and covariant derivatives from the previous two chapters. The discussion here will lead us to an operator equation for the curvature tensor that is very useful in actually computing the curvature for different spacetimes. 1 1 ^(1){ }^{1}1

35.1 Geodesic deviation (slight return)Chapter summary372
Exercises ..... 372
\checkmark This chapter revisits the Riemann curvature tensor using the geometrical tools we have built in this part of the book, allowing us to rederive book, allowing us to rederive
some key results in a more elegant manner and derive some new ones. This material pronew ones. This material pro-
vides an insight into the role vides an insight into the role
curvature plays in the physics curvature plays in the physics of the Universe.

35.1 Geodesic deviation (slight return)

In (3+1) dimensions, the Riemann tensor R ( , , R ( , , R(,,\boldsymbol{R}(,,R(,,, ) i s a ( 1 , 3 ) ) i s a ( 1 , 3 ) )isa(1,3)) is a (1,3))isa(1,3) object that encodes the curvature of spacetime. If we insert three carefully chosen vectors into the latter slots, R R R\boldsymbol{R}R returns a vector whose physical interpretation is that it encodes the relative acceleration of neighbouring geodesics. The properties of the tensor can be derived by considering the deviation of two freely falling particles following separate geodesics, much as in the thought experiment described above. This is our first task in this chapter.
Fig. 35.1 Geodesics for different freely falling particles used to compute geodesic deviation.
2 2 ^(2){ }^{2}2 This expression is sometimes called the 'geodesic deviation equation', and sometimes called the 'Jacobi equation', reflecting Jacobi's discovery of it in 1837.
The separation of the geodesics is described by the vector n n n\boldsymbol{n}n. This important vector can be understood with reference to Fig. 35.1. A typical geodesic, labelled n n nnn, is parametrized with affine parameter λ λ lambda\lambdaλ and has velocity given by the value of the tangent vector field u u u\boldsymbol{u}u evaluated at a point on the geodesic. There will be lots of other geodesics in the vicinity of a particular point λ 0 λ 0 lambda_(0)\lambda_{0}λ0. The closest is labelled n + d n n + d n n+dnn+\mathrm{d} nn+dn, the next n + 2 d n n + 2 d n n+2dnn+2 \mathrm{~d} nn+2 dn and so on. This gives us access to the vector n = d / d n n = d / d n n=d//dnn=\mathrm{d} / \mathrm{d} nn=d/dn that we interpret as the providing a measure of the relative separation of neighbouring geodesics.
Now that we have the separation vector n n n\boldsymbol{n}n we need to evaluate it as the particles fall along their geodesic world lines. The tangents to the particle's geodesics are also provided by the velocity field u u u\boldsymbol{u}u evaluated at the points of interest on the relevant geodesic. (Recall that the geodesics form a congruence of curves for the vector field u u u\boldsymbol{u}u.) The falling of the particles can then be viewed in terms of the vector n n n\boldsymbol{n}n being carried along the streamlines of the velocity field u u u\boldsymbol{u}u. The change in n n n\boldsymbol{n}n is therefore described by the Lie derivative £ u n £ u n £_(u)n£_{\boldsymbol{u}} \boldsymbol{n}£un.
Now recall the notion of a field being Lie dragged. This occurs if the vector that stretches between two curves of a congruence still stretches between them after being carried along the congruence. This is exactly the situation we require here to describe the defining property of n n n\boldsymbol{n}n. From the definition of n n n\boldsymbol{n}n being the vector that links the two particles at all points as they fall, we must have the condition that, although n n n\boldsymbol{n}n can change as it the particles fall, its Lie derivative must vanish: £ u n = 0 £ u n = 0 £_(u)n=0£_{\boldsymbol{u}} \boldsymbol{n}=0£un=0. This encodes the condition that the vector n n n\boldsymbol{n}n still stretches between the particles after being transported through u u u\boldsymbol{u}u. We are now almost ready to derive the most important equation of this chapter. This is the equation that says that relative acceleration of the geodesics is related to the Riemann tensor by 2 2 ^(2){ }^{2}2
(35.1) D 2 n d λ 2 + R ( , u , n , u ) = 0 (35.1) D 2 n d λ 2 + R ( , u , n , u ) = 0 {:(35.1)(D^(2)n)/(dlambda^(2))+R(","u","n","u)=0:}\begin{equation*} \frac{D^{2} \boldsymbol{n}}{d \lambda^{2}}+\boldsymbol{R}(, \boldsymbol{u}, \boldsymbol{n}, \boldsymbol{u})=0 \tag{35.1} \end{equation*}(35.1)D2ndλ2+R(,u,n,u)=0
In components, this reads
(35.2) D 2 n μ d λ 2 + R ν α β μ u ν n α u β = 0 (35.2) D 2 n μ d λ 2 + R ν α β μ u ν n α u β = 0 {:(35.2)(D^(2)n^(mu))/(dlambda^(2))+R_(nu alpha beta)^(mu)u^(nu)n^(alpha)u^(beta)=0:}\begin{equation*} \frac{D^{2} n^{\mu}}{d \lambda^{2}}+R_{\nu \alpha \beta}^{\mu} u^{\nu} n^{\alpha} u^{\beta}=0 \tag{35.2} \end{equation*}(35.2)D2nμdλ2+Rναβμuνnαuβ=0
We shall provide the geometric derivation of this equation and this will also give us the Riemann tensor.

Example 35.1

Before we get there, it's very useful, as a warm up, to first consider the behaviour of two particles falling in a gravitational field according to Newtonian gravitation. Specifically, let's consider the vector n n nnn, with components n k n k n^(k)n^{k}nk, that connects the trajectories of the two particles. We'll calculate the relative acceleration n ¨ k n ¨ k n^(¨)^(k)\ddot{n}^{k}n¨k. This is useful as it closely mirrors the full relativistic derivation that we shall discuss afterwards. The vector of interest is n d / d n = ( d x k / d n ) ( / x k ) n d / d n = d x k / d n / x k n-=d//dn=(dx^(k)//dn)(del//delx^(k))\boldsymbol{n} \equiv \mathrm{d} / \mathrm{d} n=\left(\mathrm{d} x^{k} / \mathrm{d} n\right)\left(\partial / \partial x^{k}\right)nd/dn=(dxk/dn)(/xk). The component n k n k n^(k)n^{k}nk is therefore d x k / d n d x k / d n dx^(k)//dn\mathrm{d} x^{k} / \mathrm{d} ndxk/dn. We therefore want to find
(35.3) 2 n k t 2 = 2 t 2 d x k d n = d d n 2 x k t 2 (35.3) 2 n k t 2 = 2 t 2 d x k d n = d d n 2 x k t 2 {:(35.3)(del^(2)n^(k))/(delt^(2))=(del^(2))/(delt^(2))((d)x^(k))/((d)n)=(d)/((d)n)(del^(2)x^(k))/(delt^(2)):}\begin{equation*} \frac{\partial^{2} n^{k}}{\partial t^{2}}=\frac{\partial^{2}}{\partial t^{2}} \frac{\mathrm{~d} x^{k}}{\mathrm{~d} n}=\frac{\mathrm{d}}{\mathrm{~d} n} \frac{\partial^{2} x^{k}}{\partial t^{2}} \tag{35.3} \end{equation*}(35.3)2nkt2=2t2 dxk dn=d dn2xkt2
The acceleration of a particle at coordinate x k x k x^(k)x^{k}xk is given by Newton's force law
(35.4) 2 x k t 2 + η i k Φ x i = 0 (35.4) 2 x k t 2 + η i k Φ x i = 0 {:(35.4)(del^(2)x^(k))/(delt^(2))+eta^(ik)(del Phi)/(delx^(i))=0:}\begin{equation*} \frac{\partial^{2} x^{k}}{\partial t^{2}}+\eta^{i k} \frac{\partial \Phi}{\partial x^{i}}=0 \tag{35.4} \end{equation*}(35.4)2xkt2+ηikΦxi=0
where Φ Φ Phi\PhiΦ is the Newtonian potential, and so we have
2 n k t 2 = d d n 2 x k t 2 (from eqn 35.3) = η i k d dn Φ x i (substituting eqn 35.4) = η i k d x j dn x j Φ x i (chain rule) = η i k 2 Φ x j x i n j = R k j n j (tidying). 2 n k t 2 = d d n 2 x k t 2  (from eqn 35.3)  = η i k d dn Φ x i  (substituting eqn 35.4)  = η i k d x j dn x j Φ x i  (chain rule)  = η i k 2 Φ x j x i n j = R k j n j  (tidying).  {:[(del^(2)n^(k))/(delt^(2))=(d)/((d)n)(del^(2)x^(k))/(delt^(2))" (from eqn 35.3) "],[=-eta^(ik)((d))/(dn)(del Phi)/(delx^(i))" (substituting eqn 35.4) "],[=-eta^(ik)((d)x^(j))/(dn)*(del)/(delx^(j))*(del Phi)/(delx^(i))" (chain rule) "],[=-eta^(ik)(del^(2)Phi)/(delx^(j)delx^(i))*n^(j)=-R^(k)_(j)n^(j)" (tidying). "]:}\begin{aligned} \frac{\partial^{2} n^{k}}{\partial t^{2}} & =\frac{\mathrm{d}}{\mathrm{~d} n} \frac{\partial^{2} x^{k}}{\partial t^{2}} & & \text { (from eqn 35.3) } \\ & =-\eta^{i k} \frac{\mathrm{~d}}{\mathrm{dn}} \frac{\partial \Phi}{\partial x^{i}} & & \text { (substituting eqn 35.4) } \\ & =-\eta^{i k} \frac{\mathrm{~d} x^{j}}{\mathrm{dn}} \cdot \frac{\partial}{\partial x^{j}} \cdot \frac{\partial \Phi}{\partial x^{i}} & & \text { (chain rule) } \\ & =-\eta^{i k} \frac{\partial^{2} \Phi}{\partial x^{j} \partial x^{i}} \cdot n^{j}=-R^{k}{ }_{j} n^{j} & & \text { (tidying). } \end{aligned}2nkt2=d dn2xkt2 (from eqn 35.3) =ηik ddnΦxi (substituting eqn 35.4) =ηik dxjdnxjΦxi (chain rule) =ηik2Φxjxinj=Rkjnj (tidying). 
This gives us an equation of motion n ¨ k + R k j n j = 0 n ¨ k + R k j n j = 0 n^(¨)^(k)+R^(k)_(j)n^(j)=0\ddot{n}^{k}+R^{k}{ }_{j} n^{j}=0n¨k+Rkjnj=0, and an expression for the components 3 R k j 3 R k j ^(3)R^(k)_(j){ }^{3} R^{k}{ }_{j}3Rkj of a tensor that can be used to determine the acceleration of the separation of the trajectories.
Let's now consider the full, geometrical version of the problem considered in the previous example. This follows exactly the same pattern: we simply want to calculate the acceleration of n n n\boldsymbol{n}n. That is to say, we want to find the double derivative of n n n\boldsymbol{n}n with respect to an affine parameter: D 2 / d λ 2 D 2 / d λ 2 D^(2)//dlambda^(2)\mathrm{D}^{2} / \mathrm{d} \lambda^{2}D2/dλ2, which is equivalent to u u n u u n grad_(u)grad_(u)n\nabla_{\boldsymbol{u}} \nabla_{\boldsymbol{u}} \boldsymbol{n}uun. From above, we recall that the relative separation vector n n n\boldsymbol{n}n is Lie dragged, which is to say that
(35.6) £ u n = [ u , n ] = 0 (35.6) £ u n = [ u , n ] = 0 {:(35.6)£_(u)n=[u","n]=0:}\begin{equation*} £_{\boldsymbol{u}} \boldsymbol{n}=[\boldsymbol{u}, \boldsymbol{n}]=0 \tag{35.6} \end{equation*}(35.6)£un=[u,n]=0
We also know that in a torsion-free system u n n u = [ u , n ] u n n u = [ u , n ] grad_(u)n-grad_(n)u=[u,n]\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}-\boldsymbol{\nabla}_{\boldsymbol{n}} \boldsymbol{u}=[\boldsymbol{u}, \boldsymbol{n}]unnu=[u,n], so the Lie dragging is equivalently described by the equation
(35.7) u n = n u (35.7) u n = n u {:(35.7)grad_(u)n=grad_(n)u:}\begin{equation*} \nabla_{u} n=\nabla_{n} u \tag{35.7} \end{equation*}(35.7)un=nu
which is the symmetry of the covariant derivative, from the last chapter. We now have the tools at our disposal to find u u n u u n grad_(u)grad_(u)n\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}uun.
Example 3 5 . 2 3 5 . 2 35.2\mathbf{3 5 . 2}35.2
Consider a geodesic with tangent vector u u u\boldsymbol{u}u. As usual it is described by the geodesic equation u u = 0 u u = 0 grad_(u)u=0\nabla_{u} \boldsymbol{u}=0uu=0. We take the covariant derivative of this expression along the n n n\boldsymbol{n}n direction
(35.8) n u u = 0 (35.8) n u u = 0 {:(35.8)grad_(n)grad_(u)u=0:}\begin{equation*} \nabla_{n} \nabla_{u} u=0 \tag{35.8} \end{equation*}(35.8)nuu=0
Next, we use the commutator [ n , u ] = n u u n n , u = n u u n [grad_(n),grad_(u)]=grad_(n)grad_(u)-grad_(u)grad_(n)\left[\nabla_{\boldsymbol{n}}, \boldsymbol{\nabla}_{u}\right]=\boldsymbol{\nabla}_{\boldsymbol{n}} \boldsymbol{\nabla}_{\boldsymbol{u}}-\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{n}}[n,u]=nuun, to write
(35.9) ( u n + [ n , u ] ) u = 0 . (35.9) u n + n , u u = 0 . {:(35.9)(grad_(u)grad_(n)+[grad_(n),grad_(u)])u=0.:}\begin{equation*} \left(\boldsymbol{\nabla}_{u} \boldsymbol{\nabla}_{n}+\left[\nabla_{n}, \boldsymbol{\nabla}_{u}\right]\right) \boldsymbol{u}=0 . \tag{35.9} \end{equation*}(35.9)(un+[n,u])u=0.
Finally, use the symmetry of the covariant derivative n u = u n n u = u n grad_(n)u=grad_(u)n\boldsymbol{\nabla}_{\boldsymbol{n}} \boldsymbol{u}=\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{n}nu=un to say
(35.10) u u n + [ n , u ] u = 0 (35.10) u u n + n , u u = 0 {:(35.10)grad_(u)grad_(u)n+[grad_(n),grad_(u)]u=0:}\begin{equation*} \nabla_{u} \nabla_{u} n+\left[\nabla_{n}, \nabla_{u}\right] u=0 \tag{35.10} \end{equation*}(35.10)uun+[n,u]u=0
or, parametrizing the geodesic with affine parameter λ λ lambda\lambdaλ,
(35.11) D 2 n d λ 2 + [ n , u ] u = 0 (35.11) D 2 n d λ 2 + n , u u = 0 {:(35.11)(D^(2)n)/((d)lambda^(2))+[grad_(n),grad_(u)]u=0:}\begin{equation*} \frac{D^{2} \boldsymbol{n}}{\mathrm{~d} \lambda^{2}}+\left[\boldsymbol{\nabla}_{\boldsymbol{n}}, \boldsymbol{\nabla}_{\boldsymbol{u}}\right] \boldsymbol{u}=0 \tag{35.11} \end{equation*}(35.11)D2n dλ2+[n,u]u=0
Although this looks like it solves the problem via an operator [ n , u n , u [grad_(n),grad_(u):}\left[\boldsymbol{\nabla}_{\boldsymbol{n}}, \boldsymbol{\nabla}_{\boldsymbol{u}}\right.[n,u ], it does not quite. We discuss this below.
4 4 ^(4){ }^{4}4 We define the second derivative μ ν c μ ν c grad_(mu)grad_(nu)c\boldsymbol{\nabla}_{\mu} \boldsymbol{\nabla}_{\nu} \boldsymbol{c}μνc to be μ ( ν c ) μ ν c grad_(mu)(grad_(nu)c)\boldsymbol{\nabla}_{\mu}\left(\boldsymbol{\nabla}_{\nu} \boldsymbol{c}\right)μ(νc) with components c α ; ν μ = ( c α ; ν ) ; μ c α ; ν μ = c α ; ν ; μ c^(alpha)_(;nu mu)=(c^(alpha)_(;nu))_(;mu)c^{\alpha}{ }_{; \nu \mu}=\left(c^{\alpha}{ }_{; \nu}\right)_{; \mu}cα;νμ=(cα;ν);μ.
The operator [ a , b ] a , b [grad_(a),grad_(b)]\left[\boldsymbol{\nabla}_{\boldsymbol{a}}, \boldsymbol{\nabla}_{\boldsymbol{b}}\right][a,b] isn't quite the one we need to compute the Riemann tensor R R R\boldsymbol{R}R to evaluate the curvature of general spaces. We saw above that [ u , n ] = 0 [ u , n ] = 0 [u,n]=0[\boldsymbol{u}, \boldsymbol{n}]=0[u,n]=0, which implies that a loop formed by vectors u u u\boldsymbol{u}u and n n n\boldsymbol{n}n closes. However, a loop constructed by travelling along arbitrary vectors a a a\boldsymbol{a}a then b b b\boldsymbol{b}b won't necessarily meet up with one where we travel along b b b\boldsymbol{b}b and then along a a a\boldsymbol{a}a. The distance between the end points is measured by the Lie bracket [ a , b ] [ a , b ] [a,b][\boldsymbol{a}, \boldsymbol{b}][a,b]. As a result we actually need the slightly upgraded operator [ a , b ] [ a , b ] a , b [ a , b ] [grad_(a),grad_(b)]-grad_([a,b])\left[\boldsymbol{\nabla}_{\boldsymbol{a}}, \boldsymbol{\nabla}_{\boldsymbol{b}}\right]-\boldsymbol{\nabla}_{[\boldsymbol{a}, \boldsymbol{b}]}[a,b][a,b] to correctly generalize the operator so that we capture the Riemann tensor. In a coordinate frame, we have [ e μ , e ν ] = 0 e μ , e ν = 0 [e_(mu),e_(nu)]=0\left[\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right]=0[eμ,eν]=0, simplifying the curvature operator. We shall often work in coordinate frames, making this correction a negligible detail.
The result of this argument is that the curvature tensor R R R\boldsymbol{R}R can be formed into a Riemann curvature operator R ^ R ^ hat(R)\hat{\boldsymbol{R}}R^ defined by
(35.12) R ( , c , a , b ) = R ^ ( a , b ) c = ( [ a , b ] [ a , b ] ) c (35.12) R ( , c , a , b ) = R ^ ( a , b ) c = a , b [ a , b ] c {:(35.12)R(","c","a","b)= hat(R)(a","b)c=([grad_(a),grad_(b)]-grad_([a,b]))c:}\begin{equation*} \boldsymbol{R}(, c, a, b)=\hat{\boldsymbol{R}}(\boldsymbol{a}, \boldsymbol{b}) \boldsymbol{c}=\left(\left[\nabla_{a}, \nabla_{b}\right]-\nabla_{[a, b]}\right) c \tag{35.12} \end{equation*}(35.12)R(,c,a,b)=R^(a,b)c=([a,b][a,b])c
The operator needs two vectors (here, a a a\boldsymbol{a}a and b b b\boldsymbol{b}b ) in order to build it. It can then act on a vector (here, c c c\boldsymbol{c}c ) to output another vector. In fact, as shown in the exercises, in a coordinate frame where the loop formed by the basis vectors closes, we end up with the memorable component equation 4 4 ^(4){ }^{4}4
(35.13) [ μ , ν ] c α = c ; ν μ α c ; μ ν α = R β μ ν α c β (35.13) μ , ν c α = c ; ν μ α c ; μ ν α = R β μ ν α c β {:(35.13)[grad_(mu),grad_(nu)]c^(alpha)=c_(;nu mu)^(alpha)-c_(;mu nu)^(alpha)=R_(beta mu nu)^(alpha)c^(beta):}\begin{equation*} \left[\nabla_{\mu}, \nabla_{\nu}\right] c^{\alpha}=c_{; \nu \mu}^{\alpha}-c_{; \mu \nu}^{\alpha}=R_{\beta \mu \nu}^{\alpha} c^{\beta} \tag{35.13} \end{equation*}(35.13)[μ,ν]cα=c;νμαc;μνα=Rβμναcβ
where c α c α c^(alpha)c^{\alpha}cα are the components of a vector.
The Riemann curvature operator turns out to be very useful and we shall work with it in the next chapter where we calculate the tensor for a variety of spacetimes. For now, we conclude that the Riemann curvature results from the double derivative of the vector field u u u\boldsymbol{u}u via
(35.14) u u n + R ^ ( n , u ) u = 0 (35.14) u u n + R ^ ( n , u ) u = 0 {:(35.14)grad_(u)grad_(u)n+ hat(R)(n","u)u=0:}\begin{equation*} \nabla_{u} \nabla_{u} n+\hat{R}(n, u) u=0 \tag{35.14} \end{equation*}(35.14)uun+R^(n,u)u=0
This is simply eqn 35.1 rewritten in terms of covariant derivatives and the Riemann operator. One of the most important things to note about the Riemann tensor and operator is that the curvature is represented by a double covariant derivative. This point will be important in the next chapter when we come to finding an efficient means of computing R R R\boldsymbol{R}R.

35.2 Components of the curvature tensor

Since the curvature tensor R R R\boldsymbol{R}R is a ( 1 , 3 ) ( 1 , 3 ) (1,3)(1,3)(1,3) tensor, inserting the basis vectors gives us an expression for the components R α β γ δ R α β γ δ R^(alpha)_(beta gamma delta)R^{\alpha}{ }_{\beta \gamma \delta}Rαβγδ, which can be interpreted in terms of the connection coefficients Γ α μ ν Γ α μ ν Gamma^(alpha)_(mu nu)\Gamma^{\alpha}{ }_{\mu \nu}Γαμν. To extract the components we write
(35.15) R α β γ δ = R ( ω α , e β , e γ , e δ ) = ω α , R ^ ( e γ , e δ ) e β , (35.15) R α β γ δ = R ω α , e β , e γ , e δ = ω α , R ^ e γ , e δ e β , {:(35.15)R^(alpha)_(beta gamma delta)=R(omega^(alpha),e_(beta),e_(gamma),e_(delta))=(:omega^(alpha),( hat(R))(e_(gamma),e_(delta))e_(beta):)",":}\begin{equation*} R^{\alpha}{ }_{\beta \gamma \delta}=\boldsymbol{R}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{e}_{\beta}, \boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right)=\left\langle\boldsymbol{\omega}^{\alpha}, \hat{\boldsymbol{R}}\left(\boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right) \boldsymbol{e}_{\beta}\right\rangle, \tag{35.15} \end{equation*}(35.15)Rαβγδ=R(ωα,eβ,eγ,eδ)=ωα,R^(eγ,eδ)eβ,
where we've used the curvature operator in the final expression. Let's now show that this gives the expression for components of R R R\boldsymbol{R}R that we derived in Chapter 11.
Example 35.3
We work in a coordinate basis, so that [ e α , e β ] = 0 e α , e β = 0 grad_([e_(alpha),e_(beta)])=0\nabla_{\left[e_{\alpha}, e_{\beta}\right]}=0[eα,eβ]=0. We therefore have a curvature operator
R ^ ( e γ , e δ ) e β = [ γ , δ ] e β R ^ e γ , e δ e β = γ , δ e β hat(R)(e_(gamma),e_(delta))e_(beta)=[grad_(gamma),grad_(delta)]e_(beta)\hat{\boldsymbol{R}}\left(\boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right) \boldsymbol{e}_{\beta}=\left[\boldsymbol{\nabla}_{\gamma}, \boldsymbol{\nabla}_{\delta}\right] \boldsymbol{e}_{\beta}R^(eγ,eδ)eβ=[γ,δ]eβ
= γ δ e β δ γ e β . = γ δ e β δ γ e β . =grad_(gamma)grad_(delta)e_(beta)-grad_(delta)grad_(gamma)e_(beta).=\nabla_{\gamma} \nabla_{\delta} e_{\beta}-\nabla_{\delta} \nabla_{\gamma} e_{\beta} .=γδeβδγeβ.
Recall that we defined 5 μ e ν = Γ α μ ν e α 5 μ e ν = Γ α μ ν e α ^(5)grad_(mu)e_(nu)=Gamma^(alpha)_(mu nu)e_(alpha){ }^{5} \nabla_{\mu} \boldsymbol{e}_{\nu}=\Gamma^{\alpha}{ }_{\mu \nu} \boldsymbol{e}_{\alpha}5μeν=Γαμνeα. Acting on e ν e ν e_(nu)\boldsymbol{e}_{\nu}eν with a term like α β α β grad_(alpha)grad_(beta)\nabla_{\alpha} \nabla_{\beta}αβ we obtain 6 6 ^(6){ }^{6}6
α β e ν = α ( Γ σ β ν e σ ) = ( α Γ σ β ν ) e σ + Γ σ β ν ( α e σ ) = Γ σ β ν x α e σ + Γ σ β ν Γ λ α σ e λ . α β e ν = α Γ σ β ν e σ = α Γ σ β ν e σ + Γ σ β ν α e σ = Γ σ β ν x α e σ + Γ σ β ν Γ λ α σ e λ . {:[grad_(alpha)grad_(beta)e_(nu)=grad_(alpha)(Gamma^(sigma)_(beta nu)e_(sigma))],[=(grad_(alpha)Gamma^(sigma)_(beta nu))e_(sigma)+Gamma^(sigma)_(beta nu)(grad_(alpha)e_(sigma))],[=(delGamma^(sigma)_(beta nu))/(delx^(alpha))e_(sigma)+Gamma^(sigma)_(beta nu)Gamma^(lambda)_(alpha sigma)e_(lambda).]:}\begin{align*} \boldsymbol{\nabla}_{\alpha} \boldsymbol{\nabla}_{\beta} \boldsymbol{e}_{\nu} & =\boldsymbol{\nabla}_{\alpha}\left(\Gamma^{\sigma}{ }_{\beta \nu} \boldsymbol{e}_{\sigma}\right) \\ & =\left(\boldsymbol{\nabla}_{\alpha} \Gamma^{\sigma}{ }_{\beta \nu}\right) \boldsymbol{e}_{\sigma}+\Gamma^{\sigma}{ }_{\beta \nu}\left(\boldsymbol{\nabla}_{\alpha} \boldsymbol{e}_{\sigma}\right) \\ & =\frac{\partial \Gamma^{\sigma}{ }_{\beta \nu}}{\partial x^{\alpha}} \boldsymbol{e}_{\sigma}+\Gamma^{\sigma}{ }_{\beta \nu} \Gamma^{\lambda}{ }_{\alpha \sigma} \boldsymbol{e}_{\lambda} . \end{align*}αβeν=α(Γσβνeσ)=(αΓσβν)eσ+Γσβν(αeσ)=Γσβνxαeσ+ΓσβνΓλασeλ.
Putting this together via R α β γ δ = ω α , R ^ ( e γ , e δ ) e β R α β γ δ = ω α , R ^ e γ , e δ e β R^(alpha)_(beta gamma delta)=(:omega^(alpha),( hat(R))(e_(gamma),e_(delta))e_(beta):)R^{\alpha}{ }_{\beta \gamma \delta}=\left\langle\boldsymbol{\omega}^{\alpha}, \hat{\boldsymbol{R}}\left(\boldsymbol{e}_{\gamma}, \boldsymbol{e}_{\delta}\right) \boldsymbol{e}_{\beta}\right\rangleRαβγδ=ωα,R^(eγ,eδ)eβ we have
(35.20) R α β γ δ = Γ α δ β x γ Γ α γ β x δ + Γ α γ μ Γ μ δ β Γ α δ μ Γ μ γ β , just as we found in Chapter 11. We conclude that the Riemann tensor defined (35.20) R α β γ δ = Γ α δ β x γ Γ α γ β x δ + Γ α γ μ Γ μ δ β Γ α δ μ Γ μ γ β ,  just as we found in Chapter 11. We conclude that the Riemann tensor defined  {:[(35.20)qquadR^(alpha)_(beta gamma delta)=(delGamma^(alpha)_(delta beta))/(delx^(gamma))-(delGamma^(alpha)_(gamma beta))/(delx^(delta))+Gamma^(alpha)_(gamma mu)Gamma^(mu)_(delta beta)-Gamma^(alpha)_(delta mu)Gamma^(mu)_(gamma beta)","],[" just as we found in Chapter 11. We conclude that the Riemann tensor defined "]:}\begin{align*} & \qquad R^{\alpha}{ }_{\beta \gamma \delta}=\frac{\partial \Gamma^{\alpha}{ }_{\delta \beta}}{\partial x^{\gamma}}-\frac{\partial \Gamma^{\alpha}{ }_{\gamma \beta}}{\partial x^{\delta}}+\Gamma^{\alpha}{ }_{\gamma \mu} \Gamma^{\mu}{ }_{\delta \beta}-\Gamma^{\alpha}{ }_{\delta \mu} \Gamma^{\mu}{ }_{\gamma \beta}, \tag{35.20}\\ & \text { just as we found in Chapter 11. We conclude that the Riemann tensor defined } \end{align*}(35.20)Rαβγδ=ΓαδβxγΓαγβxδ+ΓαγμΓμδβΓαδμΓμγβ, just as we found in Chapter 11. We conclude that the Riemann tensor defined  geometrically in terms of geodesic deviation is identical to the one we introduced back in Chapter 11.
The geodesic deviation equation gives us a tool to expand on the discussion of the local orthonormal frames from Chapter 10. There we introduced the freely falling frame as one example of a local inertial frame (LIF). In the next example, we shall find a useful set of coordinates that describe another sort of LIF, known as Riemann normal coordinates.

Example 35.4

This idea is most straightforwardly described in flat space. Erect a set of orthonormal axes 7 e α 7 e α ^(7)e_(alpha)^{7} \boldsymbol{e}_{\alpha}7eα at the origin and then send straight lines out in every direction. To get to any point P P P\mathcal{P}P in the neighbourhood of the origin we need only pick a straight line and then travel along it through a distance λ λ lambda\lambdaλ until we reach P P P\mathcal{P}P. The straight lines can be characterized by their (normalized, unit) gradient vectors u u u\boldsymbol{u}u [Fig. 35.2(a)]. Any point can therefore be reached by translating through a vector λ u λ u lambda u\lambda \boldsymbol{u}λu. The coordinates of this point ζ α ζ α zeta^(alpha)\zeta^{\alpha}ζα, can be written as
(35.21) ζ α e α = λ u α e α (35.21) ζ α e α = λ u α e α {:(35.21)zeta^(alpha)e_(alpha)=lambdau^(alpha)e_(alpha):}\begin{equation*} \zeta^{\alpha} \boldsymbol{e}_{\alpha}=\lambda u^{\alpha} \boldsymbol{e}_{\alpha} \tag{35.21} \end{equation*}(35.21)ζαeα=λuαeα
or ζ α = λ u α ζ α = λ u α zeta^(alpha)=lambdau^(alpha)\zeta^{\alpha}=\lambda u^{\alpha}ζα=λuα. As a concrete example, the point ( x , y ) = ( 2 , 1 ) ( x , y ) = ( 2 , 1 ) (x,y)=(2,1)(x, y)=(2,1)(x,y)=(2,1) [Fig. 35.2(b)] lies along the line defined by the tangent u = 1 5 ( 2 e x + e y ) u = 1 5 2 e x + e y u=(1)/(sqrt5)(2e_(x)+e_(y))\boldsymbol{u}=\frac{1}{\sqrt{5}}\left(2 \boldsymbol{e}_{x}+\boldsymbol{e}_{y}\right)u=15(2ex+ey), so that we have
(35.22) ζ α e α = λ 5 ( 2 e x + e y ) (35.22) ζ α e α = λ 5 2 e x + e y {:(35.22)zeta^(alpha)e_(alpha)=(lambda)/(sqrt5)(2e_(x)+e_(y)):}\begin{equation*} \zeta^{\alpha} \boldsymbol{e}_{\alpha}=\frac{\lambda}{\sqrt{5}}\left(2 \boldsymbol{e}_{x}+\boldsymbol{e}_{y}\right) \tag{35.22} \end{equation*}(35.22)ζαeα=λ5(2ex+ey)
If we simply parametrize the line using the distance, then the point ( 2 , 1 ) ( 2 , 1 ) (2,1)(2,1)(2,1) is found at λ = 5 λ = 5 lambda=sqrt5\lambda=\sqrt{5}λ=5, so the coordinates are ζ x = 2 , ζ y = 1 ζ x = 2 , ζ y = 1 zeta^(x)=2,zeta^(y)=1\zeta^{x}=2, \zeta^{y}=1ζx=2,ζy=1. If we choose our parametrization differently, these would be scaled accordingly.
The idea of Riemann normal coordinates is to generalize this procedure to curved space. So we start by erecting a set of orthonormal axes e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα at the origin. Since the generalization of a straight line is a geodesic, we send out geodesics in every direction from the origin [Fig. 35.2(c)]. The geodesics can be timelike or spacelike. They are characterized by their (unit) tangent vectors u u u\boldsymbol{u}u and are parametrized by affine parameter λ λ lambda\lambdaλ that tells us where we are on the curve. Select an event at P P P\mathcal{P}P and reach it by travelling a distance λ λ lambda\lambdaλ along the geodesic with gradient u u u\boldsymbol{u}u. The coordinates of the point P P P\mathcal{P}P are given the Riemann normal coordinates ζ μ ζ μ zeta^(mu)\zeta^{\mu}ζμ such that
ζ α e α = λ u α e α ζ α e α = λ u α e α zeta^(alpha)e_(alpha)=lambdau^(alpha)e_(alpha)\zeta^{\alpha} e_{\alpha}=\lambda u^{\alpha} e_{\alpha}ζαeα=λuαeα
(35.23)
5 5 ^(5){ }^{5}5 Alternatively, recall the definition from Chapter 34 that
(35.17) ω α , μ e ν = Γ μ ν α (35.17) ω α , μ e ν = Γ μ ν α {:(35.17)(:omega^(alpha),grad_(mu)e_(nu):)=Gamma_(mu nu)^(alpha):}\begin{equation*} \left\langle\boldsymbol{\omega}^{\alpha}, \boldsymbol{\nabla}_{\mu} \boldsymbol{e}_{\nu}\right\rangle=\Gamma_{\mu \nu}^{\alpha} \tag{35.17} \end{equation*}(35.17)ωα,μeν=Γμνα
6 6 ^(6){ }^{6}6 Reminder: Since Γ σ ν β Γ σ ν β Gamma^(sigma)_(nu beta)\Gamma^{\sigma}{ }_{\nu \beta}Γσνβ is a function, we have
(35.18) α Γ ν β σ = Γ ν β σ x α . (35.18) α Γ ν β σ = Γ ν β σ x α . {:(35.18)grad_(alpha)Gamma_(nu beta)^(sigma)=(delGamma_(nu beta)^(sigma))/(delx^(alpha)).:}\begin{equation*} \nabla_{\alpha} \Gamma_{\nu \beta}^{\sigma}=\frac{\partial \Gamma_{\nu \beta}^{\sigma}}{\partial x^{\alpha}} . \tag{35.18} \end{equation*}(35.18)αΓνβσ=Γνβσxα.
Fig. 35.2 (a) Unit tangents to geodesics in a flat plane. (b) The point ( x , y ) = ( 2 , 1 ) ( x , y ) = ( 2 , 1 ) (x,y)=(2,1)(x, y)=(2,1)(x,y)=(2,1) reached following a unit tangent by a distance given by the parameter λ λ lambda\lambdaλ. (c) Geodesics in a curved spacetime.
7 7 ^(7){ }^{7}7 We won't give the indices hats in this example to save on clutter.
9 9 ^(9){ }^{9}9 Why do we do this? Referring to Fig. 35.1 and using eqn 35.7 we could interpret the first term in eqn 35.14, in the form u u n Δ a Δ b u u n Δ a Δ b grad_(u)grad_(u)n Delta a Delta b\nabla_{u} \nabla_{u} \boldsymbol{n} \Delta a \Delta buunΔaΔb, as equivalent to the change δ u δ u delta u\delta \boldsymbol{u}δu due to parallel trans port of u u uuu around a loop with sides n Δ a n Δ a n Delta an \Delta anΔa port of u u uuu aroud we know from ides n Δ n Δ n Deltan \DeltanΔ and u Δ b u Δ b u Delta b\boldsymbol{u} \Delta buΔb, and we know from eqn 35.14 that δ u + R ^ ( n , u ) u Δ a Δ b = 0 δ u + R ^ ( n , u ) u Δ a Δ b = 0 delta u+ hat(R)(n,u)u Delta a Delta b=0\delta \boldsymbol{u}+\hat{\boldsymbol{R}}(\boldsymbol{n}, \boldsymbol{u}) \boldsymbol{u} \Delta a \Delta b=0δu+R^(n,u)uΔaΔb=0. Here
we extend this idea to describe paralwe extend this idea to describe paral-
lel transport of vector A A A\boldsymbol{A}A around a loop lel transport of vector A A A\boldsymbol{A}A around a loop
with sides u Δ a u Δ a u Delta a\boldsymbol{u} \Delta auΔa and v Δ b v Δ b v Delta b\boldsymbol{v} \Delta bvΔb to obtain the expression δ A + R ^ ( u , v ) A Δ a Δ b = 0 δ A + R ^ ( u , v ) A Δ a Δ b = 0 delta A+ hat(R)(u,v)A Delta a Delta b=0\delta \boldsymbol{A}+\hat{\boldsymbol{R}}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{A} \Delta a \Delta b=0δA+R^(u,v)AΔaΔb=0.
Fig. 35.3 Parallel transport of a vector around a loop gives access to the Riemann tensor. The vector is A A A\boldsymbol{A}A origi nally and changes to A = A + δ A A = A + δ A A^(')=A+delta A\boldsymbol{A}^{\prime}=\boldsymbol{A}+\delta \boldsymbol{A}A=A+δA after transport.
so that we have ζ α = λ u α ζ α = λ u α zeta^(alpha)=lambdau^(alpha)\zeta^{\alpha}=\lambda u^{\alpha}ζα=λuα.
In terms of the Riemann normal coordinates, the spacetime near the origin O O O\mathcal{O}O can be shown 8 8 ^(8){ }^{8}8 to have have metric components
(35.24) g μ ν = η μ ν 1 3 R μ α ν β ( O ) ζ α ζ β + O ( ζ 3 ) (35.24) g μ ν = η μ ν 1 3 R μ α ν β ( O ) ζ α ζ β + O ζ 3 {:(35.24)g_(mu nu)=eta_(mu nu)-(1)/(3)R_(mu alpha nu beta)(O)zeta^(alpha)zeta^(beta)+O(zeta^(3)):}\begin{equation*} g_{\mu \nu}=\eta_{\mu \nu}-\frac{1}{3} R_{\mu \alpha \nu \beta}(\mathcal{O}) \zeta^{\alpha} \zeta^{\beta}+O\left(\zeta^{3}\right) \tag{35.24} \end{equation*}(35.24)gμν=ημν13Rμανβ(O)ζαζβ+O(ζ3)
At the origin we have ζ α = 0 ζ α = 0 zeta^(alpha)=0\zeta^{\alpha}=0ζα=0, and so g μ ν = η μ ν g μ ν = η μ ν g_(mu nu)=eta_(mu nu)g_{\mu \nu}=\eta_{\mu \nu}gμν=ημν. Since there are no linear terms in ζ μ ζ μ zeta^(mu)\zeta^{\mu}ζμ in eqn 35.24 we have g μ ν , α ( O ) = 0 g μ ν , α ( O ) = 0 g_(mu nu,alpha)(O)=0g_{\mu \nu, \alpha}(\mathcal{O})=0gμν,α(O)=0, implying that the connection coefficients vanish. We have therefore described a LIF: flat with vanishing connection coefficients.

35.3 Parallel transport again

Using the geometrical concepts of the previous few chapters, we can return to the parallel transport method that we employed back in Chapter 11 to compute the components of the Riemann tensor. There we carried out a rather messy computation that we can simplify enormously with our new machinery. As we saw in Chapter 11, the Riemann tensor measures the curvature via the angle through which a vector changes when it is parallel transported around a closed path. Here we shall use a coordinate-free approach employing the covariant derivative to assess the change δ A δ A delta A\delta \boldsymbol{A}δA in a vector A A A\boldsymbol{A}A, when transported around the loop shown in Fig. 35.3. 9 9 ^(9){ }^{9}9
We carry a vector A A A\boldsymbol{A}A along a vector u u u\boldsymbol{u}u using the covariant derivative u A u A grad_(u)A\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{A}uA, which computes the value of A A A\boldsymbol{A}A at the tip of u u u\boldsymbol{u}u minus the value at the base. There are two contributions to this directional derivative: the total change in the vector field A A A\boldsymbol{A}A between the start and end points, and the correction due to the change in the basis vectors. The latter is achieved through subtracting off the vector A A A\boldsymbol{A}A parallel transported along the path. Since we are going to close the loop, the first of these contributions is zero (since the field A A A\boldsymbol{A}A is single valued) and we are simply left with the parallel transport correction. As a result, computing the covariant derivative of the vector field A A A\boldsymbol{A}A around a closed loop, outputs (minus) the result of parallel transporting the vector A A A\boldsymbol{A}A around the loop. Referring to Fig. 35.3, we shall show in the next example, that the change in the vector field δ A δ A delta A\delta \boldsymbol{A}δA on traversing the loop is given by
(35.25) δ A + R ^ ( u , v ) A Δ a Δ b = 0 (35.25) δ A + R ^ ( u , v ) A Δ a Δ b = 0 {:(35.25)delta A+ hat(R)(u","v)A Delta a Delta b=0:}\begin{equation*} \delta \boldsymbol{A}+\hat{\boldsymbol{R}}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{A} \Delta a \Delta b=0 \tag{35.25} \end{equation*}(35.25)δA+R^(u,v)AΔaΔb=0
where R ^ R ^ hat(R)\hat{\boldsymbol{R}}R^ is the curvature operator.

Example 35.5

In terms of the covariant derivative, the change δ A δ A -delta A-\delta \boldsymbol{A}δA is given by traversing the loop in Fig. 35.3 in an anticlockwise direction, which we reorder into contributions:
δ A = δ A = -delta A=-\delta \boldsymbol{A}=δA= + v A Δ b + v A Δ b +grad_(v)A Delta b+\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{A} \Delta b+vAΔb (moving along path labelled v Δ b v Δ b v Delta b\boldsymbol{v} \Delta bvΔb )
v A Δ b v A Δ b -grad_(v)A Delta b-\boldsymbol{\nabla}_{v} \boldsymbol{A} \Delta bvAΔb (moving along v Δ b v Δ b -v Delta b-\boldsymbol{v} \Delta bvΔb )
u A Δ a u A Δ a -grad_(u)A Delta a-\boldsymbol{\nabla}_{u} \boldsymbol{A} \Delta auAΔa (moving along u Δ a u Δ a -u Delta a-\boldsymbol{u} \Delta auΔa )
+ u A Δ a + u A Δ a +grad_(u)A Delta a+\boldsymbol{\nabla}_{u} \boldsymbol{A} \Delta a+uAΔa (moving along u Δ a u Δ a u Delta a\boldsymbol{u} \Delta auΔa )
[ u , v ] A Δ a Δ b [ u , v ] A Δ a Δ b -grad_([u,v])A Delta a Delta b-\boldsymbol{\nabla}_{[\boldsymbol{u}, \boldsymbol{v}]} \boldsymbol{A} \Delta a \Delta b[u,v]AΔaΔb (moving along [ u , v ] Δ a Δ b [ u , v ] Δ a Δ b -[u,v]Delta a Delta b-[\boldsymbol{u}, \boldsymbol{v}] \Delta a \Delta b[u,v]ΔaΔb ).
-delta A= +grad_(v)A Delta b (moving along path labelled v Delta b ) -grad_(v)A Delta b (moving along -v Delta b ) -grad_(u)A Delta a (moving along -u Delta a ) +grad_(u)A Delta a (moving along u Delta a ) -grad_([u,v])A Delta a Delta b (moving along -[u,v]Delta a Delta b ). | $-\delta \boldsymbol{A}=$ | $+\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{A} \Delta b$ | | (moving along path labelled $\boldsymbol{v} \Delta b$ ) | | ---: | :--- | ---: | :--- | | | $-\boldsymbol{\nabla}_{v} \boldsymbol{A} \Delta b$ | | (moving along $-\boldsymbol{v} \Delta b$ ) | | | $-\boldsymbol{\nabla}_{u} \boldsymbol{A} \Delta a$ | (moving along $-\boldsymbol{u} \Delta a$ ) | | | | $+\boldsymbol{\nabla}_{u} \boldsymbol{A} \Delta a$ | | (moving along $\boldsymbol{u} \Delta a$ ) | | | $-\boldsymbol{\nabla}_{[\boldsymbol{u}, \boldsymbol{v}]} \boldsymbol{A} \Delta a \Delta b$ | (moving along $-[\boldsymbol{u}, \boldsymbol{v}] \Delta a \Delta b$ ). | |
The key here is to recognize that the movement along v Δ b v Δ b v Delta b\boldsymbol{v} \Delta bvΔb is displaced from the movement along v Δ b v Δ b -v Delta b-\boldsymbol{v} \Delta bvΔb by a distance u Δ a u Δ a u Delta a\boldsymbol{u} \Delta auΔa. We can therefore make the replacement
(35.27) v A Δ b | 0 v A Δ b | u Δ a = u v A Δ a Δ b (35.27) v A Δ b 0 v A Δ b u Δ a = u v A Δ a Δ b {:(35.27)grad_(v)A Delta b|_(0)-grad_(v)A Delta b|_(u Delta a)=grad_(u)grad_(v)A Delta a Delta b:}\begin{equation*} \left.\nabla_{v} A \Delta b\right|_{0}-\left.\nabla_{v} A \Delta b\right|_{u \Delta a}=\nabla_{u} \nabla_{v} A \Delta a \Delta b \tag{35.27} \end{equation*}(35.27)vAΔb|0vAΔb|uΔa=uvAΔaΔb
Making an analogous replacement for the displacements along u Δ a u Δ a u Delta a\boldsymbol{u} \Delta auΔa we end up with
δ A = ( u v v u [ u , v ] ) A Δ a Δ b (35.28) = R ^ ( u , v ) A Δ a Δ b . δ A = u v v u [ u , v ] A Δ a Δ b (35.28) = R ^ ( u , v ) A Δ a Δ b . {:[-delta A=(grad_(u)grad_(v)-grad_(v)grad_(u)-grad_([u,v]))A Delta a Delta b],[(35.28)= hat(R)(u","v)A Delta a Delta b.]:}\begin{align*} -\delta \boldsymbol{A} & =\left(\boldsymbol{\nabla}_{\boldsymbol{u}} \boldsymbol{\nabla}_{\boldsymbol{v}}-\boldsymbol{\nabla}_{\boldsymbol{v}} \boldsymbol{\nabla}_{\boldsymbol{u}}-\boldsymbol{\nabla}_{[\boldsymbol{u}, \boldsymbol{v}]}\right) \boldsymbol{A} \Delta a \Delta b \\ & =\hat{\boldsymbol{R}}(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{A} \Delta a \Delta b . \tag{35.28} \end{align*}δA=(uvvu[u,v])AΔaΔb(35.28)=R^(u,v)AΔaΔb.
The Riemann operator is, therefore,
R ^ ( u , v ) = u v v u [ u , v ] ) (35.29) = [ u , v ] [ u , v ] . R ^ ( u , v ) = u v v u [ u , v ] (35.29) = u , v [ u , v ] . {:[ hat(R)(u","v){:=grad_(u)grad_(v)-grad_(v)grad_(u)-grad_([u,v]))],[(35.29)=[grad_(u),grad_(v)]-grad_([u,v]).]:}\begin{align*} \hat{R}(u, v) & \left.=\nabla_{u} \nabla_{v}-\nabla_{v} \nabla_{u}-\nabla_{[u, v]}\right) \\ & =\left[\nabla_{u}, \nabla_{v}\right]-\nabla_{[u, v]} . \tag{35.29} \end{align*}R^(u,v)=uvvu[u,v])(35.29)=[u,v][u,v].
This is, of course, exactly what we had in the previous section.
This general discussion of parallel transport gives us yet another insight into what curvature is. If we take two vectors that are orthogonal at a point and parallel transport the pair, then they must remain orthogonal, by definition of parallel transport. This seems like a way to construct an orthogonal Cartesian grid across all spacetime. However, it is just an illusion. Parallel transport is defined along a particular curve, and we have just seen how transporting a vector around a loop of curves leads, in curved space, to the vector rotating. This means that the vector field created through the parallel transport of a vector defined at some point P P P\mathcal{P}P is not single valued. (This is because the vector rotates, giving two vectors when we carry out a loop starting at P P P\mathcal{P}P : the original one and the rotated one.) We can therefore conclude that another description of curvature is the property that prevents us using parallel transport to set up a Cartesian grid across all spacetime.
Finally, in our discussion of the properties of R R R\boldsymbol{R}R, we consider how the symmetries affect the number of independent components that the Riemann tensor carries. 10 10 ^(10){ }^{10}10

Example 35.6

How many independent components does R α β γ δ R α β γ δ R_(alpha beta gamma delta)R_{\alpha \beta \gamma \delta}Rαβγδ have in n n nnn dimensions? Solution: The number of components is
(35.32) ( Number of ways of choosing α β γ δ subject to pair symmetries ) ( Number of constraints due to cyclic symmetry ) . (35.32)  Number of ways of   choosing  α β γ δ  subject   to pair symmetries  (  Number of constraints   due to cyclic symmetry  ) {:(35.32)([" Number of ways of "],[" choosing "alpha beta gamma delta" subject "],[" to pair symmetries "])-((" Number of constraints ")/(" due to cyclic symmetry "))". ":}\left(\begin{array}{c} \text { Number of ways of } \tag{35.32}\\ \text { choosing } \alpha \beta \gamma \delta \text { subject } \\ \text { to pair symmetries } \end{array}\right)-\binom{\text { Number of constraints }}{\text { due to cyclic symmetry }} \text {. }(35.32)( Number of ways of  choosing αβγδ subject  to pair symmetries )( Number of constraints  due to cyclic symmetry )
To understand the first term, we note that the antisymmetry of ( α β ) ( α β ) (alpha beta)(\alpha \beta)(αβ) and ( γ δ ) ( γ δ ) (gamma delta)(\gamma \delta)(γδ) means there are M = n 2 ( n 1 ) M = n 2 ( n 1 ) M=(n)/(2)(n-1)M=\frac{n}{2}(n-1)M=n2(n1) ways of choosing the pairs ( α β ) ( α β ) (alpha beta)(\alpha \beta)(αβ) and M M MMM of choosing ( γ δ ) ( γ δ ) (gamma delta)(\gamma \delta)(γδ). The symmetry with respect to exchanging the pairs means there are M ( M + 1 ) / 2 M ( M + 1 ) / 2 M(M+1)//2M(M+1) / 2M(M+1)/2 independent choices of pairs or pairs (i.e. a M M MMM-dimensional symmetric matrix A i j A i j A_(ij)A_{i j}Aij has M M MMM diagonal components and ( M 1 ) / 2 ( M 1 ) / 2 (M-1)//2(M-1) / 2(M1)/2 independent off-diagonal ones). We conclude that M 2 ( M + 1 ) M 2 ( M + 1 ) (M)/(2)(M+1)\frac{M}{2}(M+1)M2(M+1) ways of choosing α β γ δ α β γ δ alpha beta gamma delta\alpha \beta \gamma \deltaαβγδ, when we take all of the pair symmetries into account, where M = n ( n 1 ) / 2 M = n ( n 1 ) / 2 M=n(n-1)//2M=n(n-1) / 2M=n(n1)/2.
To understand the second term, we start with the cyclic symmetry which is encoded in the expression
(35.33) R α β γ δ + R α δ β γ + R α γ δ β = 0 (35.33) R α β γ δ + R α δ β γ + R α γ δ β = 0 {:(35.33)R_(alpha beta gamma delta)+R_(alpha delta beta gamma)+R_(alpha gamma delta beta)=0:}\begin{equation*} R_{\alpha \beta \gamma \delta}+R_{\alpha \delta \beta \gamma}+R_{\alpha \gamma \delta \beta}=0 \tag{35.33} \end{equation*}(35.33)Rαβγδ+Rαδβγ+Rαγδβ=0
10 10 ^(10){ }^{10}10 Recall the discussion of Chapter 11, where we saw that the Riemann tensor obeyed the pair-symmetry equations
R α β γ δ = R β α γ δ , R α β γ δ = R α β δ γ , (35.30) R α β γ δ = + R γ δ α β , R α β γ δ = R β α γ δ , R α β γ δ = R α β δ γ , (35.30) R α β γ δ = + R γ δ α β , {:[R_(alpha beta gamma delta)=-R_(beta alpha gamma delta)","],[R_(alpha beta gamma delta)=-R_(alpha beta delta gamma)","],[(35.30)R_(alpha beta gamma delta)=+R_(gamma delta alpha beta)","]:}\begin{align*} & R_{\alpha \beta \gamma \delta}=-R_{\beta \alpha \gamma \delta}, \\ & R_{\alpha \beta \gamma \delta}=-R_{\alpha \beta \delta \gamma}, \\ & R_{\alpha \beta \gamma \delta}=+R_{\gamma \delta \alpha \beta}, \tag{35.30} \end{align*}Rαβγδ=Rβαγδ,Rαβγδ=Rαβδγ,(35.30)Rαβγδ=+Rγδαβ,
along with a cyclic identity
R α β γ δ + R α δ β γ + R α γ δ β = 0 . ( 35.31 ) R α β γ δ + R α δ β γ + R α γ δ β = 0 . ( 35.31 ) R_(alpha beta gamma delta)+R_(alpha delta beta gamma)+R_(alpha gamma delta beta)=0.(35.31)R_{\alpha \beta \gamma \delta}+R_{\alpha \delta \beta \gamma}+R_{\alpha \gamma \delta \beta}=0 .(35.31)Rαβγδ+Rαδβγ+Rαγδβ=0.(35.31)
11 11 ^(11){ }^{11}11 Spacetimes with constant curvature have Ricci tenors with components R μ ν = C g μ ν R μ ν = C g μ ν R_(mu nu)=Cg_(mu nu)R_{\mu \nu}=C g_{\mu \nu}Rμν=Cgμν, with C C CCC constant. Spacetimes where R μ ν = 0 R μ ν = 0 R_(mu nu)=0R_{\mu \nu}=0Rμν=0 are called Spacetimes where R μ ν = 0 R μ ν = 0 R_(mu nu)=0R_{\mu \nu}=0Rμν=0 are called
Ricci flat. Of course, this latter propRicci flat. Of course, this latter prop-
erty does not mean that all of the components of the Riemann tensor neces sarily vanish; rather that the positive parts of curvature cancel out the negative parts when the Ricci tensor is computed. In this sense, the Ricci tensor is a sort of average of the Riemann tensor.
12 12 ^(12){ }^{12}12 The Weyl curvature tensor is invariant with respect to conformal transformation. Its components vanish in Minkowski space (as might be expected) and also in the RobertsonWalker spacetimes. These spaces are therefore sometimes called conformally flat. If the early Universe resembles a Robertson-Walker spacetime then we expect the Weyl curvature to be very small, or zero. As matter clumps together and black holes are formed, the Weyl curvature must increase, diverging at the black hole singularities. However, the idea that at the initial Big-Bang singularity the Weyl curvature is constrained to be small (or zero) is known as the Weyl curvature hypothesis, and is a strong constraint on cosmological models.
This means we need all four indices to be distinct. The number of constraints is then the number of combinations of four objects can be taken from a collection of n n nnn objects, which is
(35.34) n ! ( n 4 ) ! 4 ! (35.35) M 2 ( M + 1 ) n ! ( n 4 ) ! 4 ! = n 2 ( n 2 1 ) 12 (35.34) n ! ( n 4 ) ! 4 ! (35.35) M 2 ( M + 1 ) n ! ( n 4 ) ! 4 ! = n 2 n 2 1 12 {:[(35.34)(n!)/((n-4)!4!)],[(35.35)(M)/(2)(M+1)-(n!)/((n-4)!4!)=(n^(2)(n^(2)-1))/(12)]:}\begin{gather*} \frac{n!}{(n-4)!4!} \tag{35.34}\\ \frac{M}{2}(M+1)-\frac{n!}{(n-4)!4!}=\frac{n^{2}\left(n^{2}-1\right)}{12} \tag{35.35} \end{gather*}(35.34)n!(n4)!4!(35.35)M2(M+1)n!(n4)!4!=n2(n21)12
or
In n = 4 n = 4 n=4n=4n=4 dimensions, we have twenty independent components.
After having derived the properties of the Riemann curvature tensor and operator, we shall, in the next chapter, finally calculate its components for a variety of spacetimes. Before we get there we pause to consider the Ricci tensor, which incorporates part of the Riemann tensor into Einstein's equation.

35.4 The meaning of the Ricci tensor

The Ricci tensor is the result of the only contraction of the Riemann tensor that does not vanish. In four spacetime dimensions, it has ten of the twenty independent components of the Riemann tensor. Physically, it can be thought of as that part of the Riemann curvature that causes volumes of matter (i.e. sources of curvature) to shrink. As a consequence of the Einstein equation, the Ricci tensor vanishes in free space. 11 11 ^(11){ }^{11}11 In contrast, if we remove the Ricci tensor part from the Riemann tensor, we obtain the Weyl curvature tensor C C C\boldsymbol{C}C, defined as having components
C μ ν α β = R μ ν α β 1 2 R μ α g ν β + 1 2 R μ β g ν α + 1 2 R ν α g μ β 1 2 R ν β g μ α (35.36) + 1 6 R ( g μ α g ν β g μ β g ν α ) C μ ν α β = R μ ν α β 1 2 R μ α g ν β + 1 2 R μ β g ν α + 1 2 R ν α g μ β 1 2 R ν β g μ α (35.36) + 1 6 R g μ α g ν β g μ β g ν α {:[C_(mu nu alpha beta)=R_(mu nu alpha beta)-(1)/(2)R_(mu alpha)g_(nu beta)+(1)/(2)R_(mu beta)g_(nu alpha)+(1)/(2)R_(nu alpha)g_(mu beta)-(1)/(2)R_(nu beta)g_(mu alpha)],[(35.36)+(1)/(6)R(g_(mu alpha)g_(nu beta)-g_(mu beta)g_(nu alpha))]:}\begin{align*} C_{\mu \nu \alpha \beta}= & R_{\mu \nu \alpha \beta}-\frac{1}{2} R_{\mu \alpha} g_{\nu \beta}+\frac{1}{2} R_{\mu \beta} g_{\nu \alpha}+\frac{1}{2} R_{\nu \alpha} g_{\mu \beta}-\frac{1}{2} R_{\nu \beta} g_{\mu \alpha} \\ & +\frac{1}{6} R\left(g_{\mu \alpha} g_{\nu \beta}-g_{\mu \beta} g_{\nu \alpha}\right) \tag{35.36} \end{align*}Cμναβ=Rμναβ12Rμαgνβ+12Rμβgνα+12Rναgμβ12Rνβgμα(35.36)+16R(gμαgνβgμβgνα)
The Weyl tensor contains the other ten independent components of R R R\boldsymbol{R}R and does not generally vanish in the vacuum, in contrast to the Ricci tensor. 12 12 ^(12){ }^{12}12 An important physical effect of the Weyl curvature tensor C C C\boldsymbol{C}C is in its role to distort geodesics.
Example 35.7
Since Ricci curvature can sometimes also distort timelike geodesics, it's best to compare the two sorts of curvature in terms of their action on light rays (or null geodesics), The trace-reversed part of the Ricci tensor has components
(35.37) G μ ν = R μ ν 1 4 R g μ ν (35.37) G μ ν = R μ ν 1 4 R g μ ν {:(35.37)G_(mu nu)=R_(mu nu)-(1)/(4)Rg_(mu nu):}\begin{equation*} G_{\mu \nu}=R_{\mu \nu}-\frac{1}{4} R g_{\mu \nu} \tag{35.37} \end{equation*}(35.37)Gμν=Rμν14Rgμν
This part of the curvature causes light rays from a source to focus, just like a positive focusing lens. The part of the curvature captured by the Weyl curvature has the effect of an astigmatic lens, which focuses positively in one plane and defocuses in a perpendicular plane.
Roger Penrose invites us to think about this 13 13 ^(13){ }^{13}13 by imagining the action of a transparent non-refracting Sun on rays from distant stars that lie behind it, as shown in Fig. 35.4. Without any gravitational field the light from each star should form a circular image, as shown by the circles drawn with dashed lines. In the presence of the gravitational field from our hypothetical transparent Sun, those rays that pass through the Sun are mostly affected by the Ricci part of the curvature, causing them to be magnified by the positive focusing effect to form a (larger) circular image (shown by the circles drawn with solid lines). On the other hand, stars whose rays pass beyond the rim of the Sun will only experience the Weyl effect, which causes their images to be distorted into ellipses (as shown in the bottom left-hand corner of Fig. 35.4).
We can make these ideas a little more mathematical by considering geodesic deviation.

Example 35.8

Recall the geodesic equation in component form
(35.38) D 2 n μ d λ 2 = ( R ν α β μ u ν u β ) n α (35.38) D 2 n μ d λ 2 = R ν α β μ u ν u β n α {:(35.38)(D^(2)n^(mu))/(dlambda^(2))=(-R_(nu alpha beta)^(mu)u^(nu)u^(beta))n^(alpha):}\begin{equation*} \frac{\mathrm{D}^{2} n^{\mu}}{\mathrm{d} \lambda^{2}}=\left(-R_{\nu \alpha \beta}^{\mu} u^{\nu} u^{\beta}\right) n^{\alpha} \tag{35.38} \end{equation*}(35.38)D2nμdλ2=(Rναβμuνuβ)nα
Notice how the velocity components are contracted against the indices in the second and fourth positions but that we trace over the other two components to make the Ricci tensor.
Consider the matrix K μ α = R μ ν α β u ν u β K μ α = R μ ν α β u ν u β K^(mu)_(alpha)=-R^(mu)_(nu alpha beta)u^(nu)u^(beta)K^{\mu}{ }_{\alpha}=-R^{\mu}{ }_{\nu \alpha \beta} u^{\nu} u^{\beta}Kμα=Rμναβuνuβ. If we choose n n nnn to be an eigenvector of this matrix, then K n = λ n K n = λ n Kn=lambda nK \boldsymbol{n}=\lambda \boldsymbol{n}Kn=λn, outputting a parallel vector that is scaled by some amount. If we take the eigenvectors of K K K\boldsymbol{K}K to describe a volume in spacetime, then the trace over K K K\boldsymbol{K}K gives us the rate of change of the volume. 14 14 ^(14){ }^{14}14 We have then that
(35.40) K μ μ = R ν μ β μ u ν u β = R ν β u ν u β (35.40) K μ μ = R ν μ β μ u ν u β = R ν β u ν u β {:(35.40)K_(mu)^(mu)=-R_(nu mu beta)^(mu)u^(nu)u^(beta)=-R_(nu beta)u^(nu)u^(beta):}\begin{equation*} K_{\mu}^{\mu}=-R_{\nu \mu \beta}^{\mu} u^{\nu} u^{\beta}=-R_{\nu \beta} u^{\nu} u^{\beta} \tag{35.40} \end{equation*}(35.40)Kμμ=Rνμβμuνuβ=Rνβuνuβ
where R ν β R ν β R_(nu beta)R_{\nu \beta}Rνβ are the components of the Ricci tensor. The rate of change of the volume element δ V δ V delta V\delta VδV is then
(35.41) D 2 δ V d λ 2 = ( R ν β u ν u β ) δ V (35.41) D 2 δ V d λ 2 = R ν β u ν u β δ V {:(35.41)(D^(2)delta V)/((d)lambda^(2))=(-R_(nu beta)u^(nu)u^(beta))delta V:}\begin{equation*} \frac{\mathrm{D}^{2} \delta V}{\mathrm{~d} \lambda^{2}}=\left(-R_{\nu \beta} u^{\nu} u^{\beta}\right) \delta V \tag{35.41} \end{equation*}(35.41)D2δV dλ2=(Rνβuνuβ)δV
This demonstrates that the Ricci tensor describes the evolution of volumes. The contraction that creates it, which involves tracing over the components of the Riemann tensor, amounts to constructing the response of the volume from the individual lengths that make it up.
Returning the to full geodesic equation D 2 n μ / d λ 2 = K μ ν n ν D 2 n μ / d λ 2 = K μ ν n ν D^(2)n^(mu)//dlambda^(2)=K^(mu)_(nu)n^(nu)\mathrm{D}^{2} n^{\mu} / \mathrm{d} \lambda^{2}=K^{\mu}{ }_{\nu} n^{\nu}D2nμ/dλ2=Kμνnν, we could also consider the fate of a small sphere as we move along a geodesic. We choose geodesics that follow the edges of the sphere to be initially parallel to the geodesic whose tangent is u u u\boldsymbol{u}u. Then geodesic deviation causes the sphere to deform into an ellipsoid, whose axes are the eigenvalues of K K K\boldsymbol{K}K.
This chapter has considered the geometric origin of the Riemann tensor, and we have already said that this tensor derives from the fundamental field of general relativity, the metric field. In the following chapter, we consider the most efficient method to extract the Riemann tensor from a metric.
13 13 ^(13){ }^{13}13 This is discussed further in Penrose (2004), Section 28.8.
Fig. 35.4 The magnification and distortion of the images of distant stars due to a transparent Sun. The stars actual sizes are shown by the dashed circles; their images are shown by solid lines.
14 14 ^(14){ }^{14}14 To see this consider a twodimensional area A = δ x δ y A = δ x δ y A=delta x delta yA=\delta x \delta yA=δxδy where δ x δ x delta x\delta xδx and δ y δ y delta y\delta yδy are the eigenfunctions of a linear operator D 2 / d λ 2 = K ^ D 2 / d λ 2 = K ^ D^(2)//dlambda^(2)= hat(K)\mathrm{D}^{2} / \mathrm{d} \lambda^{2}=\hat{K}D2/dλ2=K^. Operating on A A AAA we obtain
D 2 A / d λ 2 = K ^ A D 2 A / d λ 2 = K ^ A D^(2)A//dlambda^(2)= hat(K)A\mathrm{D}^{2} A / \mathrm{d} \lambda^{2}=\hat{K} AD2A/dλ2=K^A
= ( λ x δ x ) δ y + δ x ( λ y δ y ) (35.39) = ( λ x + λ y ) A = ( Tr K ^ ) A . = λ x δ x δ y + δ x λ y δ y (35.39) = λ x + λ y A = ( Tr K ^ ) A . {:[=(lambda_(x)delta x)delta y+delta x(lambda_(y)delta y)],[(35.39)=(lambda_(x)+lambda_(y))A=(Tr hat(K))A.]:}\begin{align*} & =\left(\lambda_{x} \delta x\right) \delta y+\delta x\left(\lambda_{y} \delta y\right) \\ & =\left(\lambda_{x}+\lambda_{y}\right) A=(\operatorname{Tr} \hat{K}) A . \tag{35.39} \end{align*}=(λxδx)δy+δx(λyδy)(35.39)=(λx+λy)A=(TrK^)A.

Chapter summary

  • Geodesic deviation can be calculated geometrically by considering the condition that the Lie derivative £ u n £ u n £_(u)n£_{u} n£un vanishes.
  • Instead of the Riemann tensor, we can use the Riemann curvature operator R ^ R ^ hat(R)\hat{\boldsymbol{R}}R^, defined by
(35.42) R ^ ( a , b ) c = ( [ a , b ] [ a , b ] ) c . (35.42) R ^ ( a , b ) c = a , b [ a , b ] c . {:(35.42) hat(R)(a","b)c=([grad_(a),grad_(b)]-grad_([a,b]))c.:}\begin{equation*} \hat{R}(a, b) c=\left(\left[\nabla_{a}, \nabla_{b}\right]-\nabla_{[a, b]}\right) c . \tag{35.42} \end{equation*}(35.42)R^(a,b)c=([a,b][a,b])c.
  • The Ricci tensor represents the part of the Riemann tensor that has a focussing effect on volumes. The Weyl tensor represents the parts left over, which have a distorting effect on light rays.

Exercises

(35.1) Show that, in terms of coordinates, [ X , Y ] Z [ X , Y ] Z grad_([X,Y])Z\boldsymbol{\nabla}_{[X, Y]} \boldsymbol{Z}[X,Y]Z can be represented as
(35.43) ( X γ Y ; γ δ Y γ X ; γ δ ) Z ; δ α (35.43) X γ Y ; γ δ Y γ X ; γ δ Z ; δ α {:(35.43)(X^(gamma)Y_(;gamma)^(delta)-Y^(gamma)X_(;gamma)^(delta))Z_(;delta)^(alpha):}\begin{equation*} \left(X^{\gamma} Y_{; \gamma}^{\delta}-Y^{\gamma} X_{; \gamma}^{\delta}\right) Z_{; \delta}^{\alpha} \tag{35.43} \end{equation*}(35.43)(XγY;γδYγX;γδ)Z;δα
(35.2) (a) Show that, in a coordinate frame,
(35.44) R μ ν α β v ν = v ; β α μ v μ ; α β . (35.44) R μ ν α β v ν = v ; β α μ v μ ; α β . {:(35.44)R^(mu)_(nu alpha beta)v^(nu)=v_(;beta alpha)^(mu)-v^(mu)_(;alpha beta).:}\begin{equation*} R^{\mu}{ }_{\nu \alpha \beta} v^{\nu}=v_{; \beta \alpha}^{\mu}-v^{\mu}{ }_{; \alpha \beta} . \tag{35.44} \end{equation*}(35.44)Rμναβvν=v;βαμvμ;αβ.
(b) Show further that
(35.45) v μ ; α μ v μ ; μ α = R α μ v μ . (35.45) v μ ; α μ v μ ; μ α = R α μ v μ . {:(35.45)v^(mu;alpha)_(mu)-v^(mu)_(;mu)^(alpha)=R^(alpha)_(mu)v^(mu).:}\begin{equation*} v^{\mu ; \alpha}{ }_{\mu}-v^{\mu}{ }_{; \mu}^{\alpha}=R^{\alpha}{ }_{\mu} v^{\mu} . \tag{35.45} \end{equation*}(35.45)vμ;αμvμ;μα=Rαμvμ.
(35.3) Consider the flat space equation
(35.46) 2 A μ + μ ν A ν = J μ (35.46) 2 A μ + μ ν A ν = J μ {:(35.46)-del^(2)A^(mu)+del^(mu)del_(nu)A^(nu)=J^(mu):}\begin{equation*} -\partial^{2} A^{\mu}+\partial^{\mu} \partial_{\nu} A^{\nu}=J^{\mu} \tag{35.46} \end{equation*}(35.46)2Aμ+μνAν=Jμ
(a) Write the equation in index comma notation. (b) Show that using the comma goes to semicolon rule results in an ambiguity.
(c) Use the result of Exercise 35.2 to show that the two possible forms of the equation can be related using the component of the Ricci tensor R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν. There is no definite rule on how to avoid this ambiguity. Misner, Thorne, and Wheeler give a set of rules of thumb to choose the most physically reasonable.
(35.4) We are now used to setting up a local orthonormal frame at some point in curved spacetime. Here, following the approach of Poisson and Will, we shall set up Riemann normal coordinates ζ μ ζ μ zeta^(mu)\zeta^{\mu}ζμ, which provide a local inertial frame where, in addition
to the property g μ ^ ν ^ ( O ) = η μ ^ ν ^ g μ ^ ν ^ ( O ) = η μ ^ ν ^ g_( hat(mu) hat(nu))(O)=eta_( hat(mu) hat(nu))g_{\hat{\mu} \hat{\nu}}(\mathcal{O})=\eta_{\hat{\mu} \hat{\nu}}gμ^ν^(O)=ημ^ν^ at the origin O O O\mathcal{O}O, the connection coefficients are also guaranteed to vanish. To do this, we shall work with the basis vectors e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ of an orthonormal frame defined at the origin. (We won't follow our usual convention of gives these directions hats in this case, to save on clutter.)
Consider the setup for the Riemann normal coordinates ζ μ = λ u μ ζ μ = λ u μ zeta^(mu)=lambdau^(mu)\zeta^{\mu}=\lambda u^{\mu}ζμ=λuμ, where u u u\boldsymbol{u}u is tangent to a geodesic and λ λ lambda\lambdaλ measures the distance from O O O\mathcal{O}O. The components here are those of an orthonormal frame that we set up at the origin O O O\mathcal{O}O (only). If we vary the directions of the tangents we obtain more geodesics that originate from O O O\mathcal{O}O. These are linked by deviation vectors n ν n ν n_(nu)\boldsymbol{n}_{\nu}nν (one for each component ν ν nu\nuν of the tangent) with components
(35.47) n ν μ = ζ μ u ν = s δ ν μ (35.47) n ν μ = ζ μ u ν = s δ ν μ {:(35.47)n_(nu)^(mu)=(delzeta^(mu))/(delu^(nu))=sdelta_(nu)^(mu):}\begin{equation*} n_{\nu}^{\mu}=\frac{\partial \zeta^{\mu}}{\partial u^{\nu}}=s \delta_{\nu}^{\mu} \tag{35.47} \end{equation*}(35.47)nνμ=ζμuν=sδνμ
For definiteness, we'll treat the geodesics as spacelike.
(a) Show that at any point on the geodesic Γ μ α β u α u β = 0 Γ μ α β u α u β = 0 Gamma^(mu)_(alpha beta)u^(alpha)u^(beta)=0\Gamma^{\mu}{ }_{\alpha \beta} u^{\alpha} u^{\beta}=0Γμαβuαuβ=0. Why does this imply that the connection coefficients vanish at the origin?
(b) Since the connection coefficients vanish at O O O\mathcal{O}O we can expand
(35.48) Γ μ α β = Γ μ α β ζ σ ζ σ + (35.48) Γ μ α β = Γ μ α β ζ σ ζ σ + {:(35.48)Gamma^(mu)_(alpha beta)=(delGamma^(mu)_(alpha beta))/(delzeta^(sigma))zeta^(sigma)+dots:}\begin{equation*} \Gamma^{\mu}{ }_{\alpha \beta}=\frac{\partial \Gamma^{\mu}{ }_{\alpha \beta}}{\partial \zeta^{\sigma}} \zeta^{\sigma}+\ldots \tag{35.48} \end{equation*}(35.48)Γμαβ=Γμαβζσζσ+
Use this to show that the first derivative of the deviation vectors is
( D n ν d λ ) μ = δ ν μ + λ 2 Γ α ν μ ( O ) ζ σ u α u σ + O ( s 3 ) D n ν d λ μ = δ ν μ + λ 2 Γ α ν μ ( O ) ζ σ u α u σ + O s 3 ((Dn_(nu))/(dlambda))^(mu)=delta_(nu)^(mu)+lambda^(2)(delGamma_(alpha nu)^(mu)(O))/(delzeta^(sigma))u^(alpha)u^(sigma)+O(s^(3))\left(\frac{D \boldsymbol{n}_{\nu}}{\mathrm{d} \lambda}\right)^{\mu}=\delta_{\nu}^{\mu}+\lambda^{2} \frac{\partial \Gamma_{\alpha \nu}^{\mu}(\mathcal{O})}{\partial \zeta^{\sigma}} u^{\alpha} u^{\sigma}+O\left(s^{3}\right)(Dnνdλ)μ=δνμ+λ2Γανμ(O)ζσuαuσ+O(s3)
(c) Show that the second derivative is
( D 2 n ν d λ 2 ) μ = 3 λ Γ α ν μ ( O ) ζ σ u α u σ + O ( s 2 ) D 2 n ν d λ 2 μ = 3 λ Γ α ν μ ( O ) ζ σ u α u σ + O s 2 ((D^(2)n_(nu))/(dlambda^(2)))^(mu)=3lambda(delGamma_(alpha nu)^(mu)(O))/(delzeta^(sigma))u^(alpha)u^(sigma)+O(s^(2))\left(\frac{D^{2} \boldsymbol{n}_{\nu}}{\mathrm{d} \lambda^{2}}\right)^{\mu}=3 \lambda \frac{\partial \Gamma_{\alpha \nu}^{\mu}(\mathcal{O})}{\partial \zeta^{\sigma}} u^{\alpha} u^{\sigma}+O\left(s^{2}\right)(D2nνdλ2)μ=3λΓανμ(O)ζσuαuσ+O(s2)
(d) Show that the geodesic deviation equation is then
(35.51) [ 3 Γ μ α ν ( O ) ζ σ + R α ν σ μ ( O ) ] u α u σ + O ( s 2 ) = 0 (35.51) 3 Γ μ α ν ( O ) ζ σ + R α ν σ μ ( O ) u α u σ + O s 2 = 0 {:(35.51)[3(delGamma^(mu)_(alpha nu)(O))/(delzeta^(sigma))+R_(alpha nu sigma)^(mu)(O)]u^(alpha)u^(sigma)+O(s^(2))=0:}\begin{equation*} \left[3 \frac{\partial \Gamma^{\mu}{ }_{\alpha \nu}(\mathcal{O})}{\partial \zeta^{\sigma}}+R_{\alpha \nu \sigma}^{\mu}(\mathcal{O})\right] u^{\alpha} u^{\sigma}+O\left(s^{2}\right)=0 \tag{35.51} \end{equation*}(35.51)[3Γμαν(O)ζσ+Rανσμ(O)]uαuσ+O(s2)=0
(e) Show that this implies that, at the origin, we have
Γ μ α ν ( O ) ζ σ + Γ μ σ ν ( O ) ζ α = 1 3 ( R α ν σ μ + R σ ν α μ ) Γ μ α ν ( O ) ζ σ + Γ μ σ ν ( O ) ζ α = 1 3 R α ν σ μ + R σ ν α μ (delGamma^(mu)_(alpha nu)(O))/(delzeta^(sigma))+(delGamma^(mu)_(sigma nu)(O))/(delzeta^(alpha))=-(1)/(3)(R_(alpha nu sigma)^(mu)+R_(sigma nu alpha)^(mu))\frac{\partial \Gamma^{\mu}{ }_{\alpha \nu}(\mathcal{O})}{\partial \zeta^{\sigma}}+\frac{\partial \Gamma^{\mu}{ }_{\sigma \nu}(\mathcal{O})}{\partial \zeta^{\alpha}}=-\frac{1}{3}\left(R_{\alpha \nu \sigma}^{\mu}+R_{\sigma \nu \alpha}^{\mu}\right)Γμαν(O)ζσ+Γμσν(O)ζα=13(Rανσμ+Rσναμ)
(f) By permuting indices, use the previous expression to obtain
(35.53) Γ ν σ μ ( O ) ζ α = 1 3 ( R ν σ α μ + R σ ν α μ ) (35.53) Γ ν σ μ ( O ) ζ α = 1 3 R ν σ α μ + R σ ν α μ {:(35.53)(delGamma_(nu sigma)^(mu)(O))/(delzeta^(alpha))=-(1)/(3)(R_(nu sigma alpha)^(mu)+R_(sigma nu alpha)^(mu)):}\begin{equation*} \frac{\partial \Gamma_{\nu \sigma}^{\mu}(\mathcal{O})}{\partial \zeta^{\alpha}}=-\frac{1}{3}\left(R_{\nu \sigma \alpha}^{\mu}+R_{\sigma \nu \alpha}^{\mu}\right) \tag{35.53} \end{equation*}(35.53)Γνσμ(O)ζα=13(Rνσαμ+Rσναμ)
(g) Use the expression from (f) to show
Hint: We saw in Exercise 34.2 that taking derivatives of g μ ν = e μ e ν g μ ν = e μ e ν g_(mu nu)=e_(mu)*e_(nu)g_{\mu \nu}=\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}gμν=eμeν yields the useful identity
(35.55) g μ ν , σ = g μ β Γ σ ν β + g ν β Γ σ μ β . (35.55) g μ ν , σ = g μ β Γ σ ν β + g ν β Γ σ μ β . {:(35.55)g_(mu nu,sigma)=g_(mu beta)Gamma_(sigma nu)^(beta)+g_(nu beta)Gamma_(sigma mu)^(beta).:}\begin{equation*} g_{\mu \nu, \sigma}=g_{\mu \beta} \Gamma_{\sigma \nu}^{\beta}+g_{\nu \beta} \Gamma_{\sigma \mu}^{\beta} . \tag{35.55} \end{equation*}(35.55)gμν,σ=gμβΓσνβ+gνβΓσμβ.
Take the derivative of this to compute the expression.
(h) Finally, expand the metric in the Riemann normal coordinates, to recover eqn 35.24 .
(35.5) Consider a two-dimensional surface spanned by basis vectors A A A\boldsymbol{A}A and B B B\boldsymbol{B}B, where [ A , B ] = 0 [ A , B ] = 0 [A,B]=0[\boldsymbol{A}, \boldsymbol{B}]=0[A,B]=0. Given that the torsion vanishes and that A A = B A A = B grad_(A)A=-B\nabla_{\boldsymbol{A}} \boldsymbol{A}=-\boldsymbol{B}AA=B, B B = B B B = B grad_(B)B=B\nabla_{B} \boldsymbol{B}=\boldsymbol{B}BB=B and B A = A B A = A grad_(B)A=A\nabla_{B} \boldsymbol{A}=\boldsymbol{A}BA=A, show that the Riemann tensor vanishes.
(35.6) Consider the case where we have the basis vectors X X X\boldsymbol{X}X and Y Y Y\boldsymbol{Y}Y, with the properties [ X , Y ] = 0 [ X , Y ] = 0 [X,Y]=0[\boldsymbol{X}, \boldsymbol{Y}]=0[X,Y]=0 and
X X = 0 , X Y = X + Y Y X = X Y , Y Y = 0 X X = 0 , X Y = X + Y Y X = X Y , Y Y = 0 {:[grad_(X)X=0",",grad_(X)Y=X+Y],[grad_(Y)X=X-Y",",grad_(Y)Y=0]:}\begin{array}{cc} \nabla_{\boldsymbol{X}} \boldsymbol{X}=0, & \nabla_{\boldsymbol{X}} \boldsymbol{Y}=\boldsymbol{X}+\boldsymbol{Y} \\ \nabla_{\boldsymbol{Y}} \boldsymbol{X}=\boldsymbol{X}-\boldsymbol{Y}, & \nabla_{Y} \boldsymbol{Y}=0 \end{array}XX=0,XY=X+YYX=XY,YY=0
(a) Compute the connection coefficients.
(b) Compute the components of the Riemann tensor.

36

36.1 Connection 1-forms 374
36.2 Two rules 377 36.3 Le repère mobile 379 379 quad379\quad 379379 36.4 Example computations 380 Chapter summary
Exercises
1 1 ^(1){ }^{1}1 Recall that we put hats on indices (e.g. A μ ^ A μ ^ A_( hat(mu))A_{\hat{\mu}}Aμ^ ) to remind ourselves of the use of the orthonormal frame.
2 2 ^(2){ }^{2}2 We can expand this using vielbein by writing g = g α β d x α d x β g = g α β d x α d x β g=g_(alpha beta)dx^(alpha)ox dx^(beta)\boldsymbol{g}=g_{\alpha \beta} \boldsymbol{d} x^{\alpha} \otimes \boldsymbol{d} x^{\beta}g=gαβdxαdxβ with components
(36.1) g α β = η μ ^ ν ^ ( e α ) μ ^ ( e β ) ν ^ (36.1) g α β = η μ ^ ν ^ e α μ ^ e β ν ^ {:(36.1)g_(alpha beta)=eta_( hat(mu) hat(nu))(e_(alpha))^( hat(mu))(e_(beta))^( hat(nu)):}\begin{equation*} g_{\alpha \beta}=\eta_{\hat{\mu} \hat{\nu}}\left(\boldsymbol{e}_{\alpha}\right)^{\hat{\mu}}\left(\boldsymbol{e}_{\beta}\right)^{\hat{\nu}} \tag{36.1} \end{equation*}(36.1)gαβ=ημ^ν^(eα)μ^(eβ)ν^
and then write orthonormal basis 1forms ω μ ^ = ( e α ) μ ^ d x α ω μ ^ = e α μ ^ d x α omega^( hat(mu))=(e_(alpha))^( hat(mu))dx^(alpha)\boldsymbol{\omega}^{\hat{\mu}}=\left(\boldsymbol{e}_{\alpha}\right)^{\hat{\mu}} \boldsymbol{d} x^{\alpha}ωμ^=(eα)μ^dxα. Putting these expressions together yields eqn 36.2.

Cartan's method

The finest collection of frames I ever saw
Sir Humphery Davy (1778-1829) when asked what he thought
of the Paris art galleries
The calculation of curvature tensors is, generally speaking, a tedious business. However, using the geometrical techniques we have built up in the last few chapters, we can come up with a more efficient means of extracting the tensor R R R\boldsymbol{R}R from the metric field tensor g g g\boldsymbol{g}g. The key, discovered by Elie Cartan, is to make use of forms. Cartan's method is especially effective because we often choose to work in the orthonormal frame 1 1 ^(1){ }^{1}1 with metric 2 2 ^(2){ }^{2}2
(36.2) d s 2 = g = η μ ^ ν ^ ω μ ^ ω ν ^ (36.2) d s 2 = g = η μ ^ ν ^ ω μ ^ ω ν ^ {:(36.2)ds^(2)=g=eta_( hat(mu) hat(nu))omega^( hat(mu))oxomega^( hat(nu)):}\begin{equation*} \boldsymbol{d} s^{2}=\boldsymbol{g}=\eta_{\hat{\mu} \hat{\nu}} \boldsymbol{\omega}^{\hat{\mu}} \otimes \boldsymbol{\omega}^{\hat{\nu}} \tag{36.2} \end{equation*}(36.2)ds2=g=ημ^ν^ωμ^ων^
where η μ ^ ν ^ η μ ^ ν ^ eta_( hat(mu) hat(nu))\eta_{\hat{\mu} \hat{\nu}}ημ^ν^ are the components of the Minkowski metric tensor. That is to say, in a typical coordinate frame ( t , χ , θ , ϕ ) ( t , χ , θ , ϕ ) (t,chi,theta,phi)(t, \chi, \theta, \phi)(t,χ,θ,ϕ) with diagonal metric, we cast the metric field in the form
(36.3) d s 2 = g = ( ω t ^ ) 2 + ( ω χ ^ ) 2 + ( ω θ ^ ) 2 + ( ω ϕ ^ ) 2 (36.3) d s 2 = g = ω t ^ 2 + ω χ ^ 2 + ω θ ^ 2 + ω ϕ ^ 2 {:(36.3)ds^(2)=g=-(omega^( hat(t)))^(2)+(omega^( hat(chi)))^(2)+(omega^( hat(theta)))^(2)+(omega^( hat(phi)))^(2):}\begin{equation*} \boldsymbol{d} s^{2}=\boldsymbol{g}=-\left(\boldsymbol{\omega}^{\hat{t}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\chi}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\theta}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\phi}}\right)^{2} \tag{36.3} \end{equation*}(36.3)ds2=g=(ωt^)2+(ωχ^)2+(ωθ^)2+(ωϕ^)2
In this chapter, we show how a routine based on taking exterior derivatives of the basis 1 -forms ω μ ^ ω μ ^ omega^( hat(mu))\boldsymbol{\omega}^{\hat{\mu}}ωμ^ does the job of extracting the curvature.

36.1 Connection 1-forms

In the last chapter, we identified a curvature operator R ^ ( a , b ) = R ^ ( a , b ) = hat(R)(a,b)=\hat{\boldsymbol{R}}(\boldsymbol{a}, \boldsymbol{b})=R^(a,b)= [ a , b ] [ a , b ] a , b [ a , b ] [grad_(a),grad_(b)]-grad_([a,b])\left[\boldsymbol{\nabla}_{\boldsymbol{a}}, \boldsymbol{\nabla}_{\boldsymbol{b}}\right]-\boldsymbol{\nabla}_{[a, b]}[a,b][a,b]. This is a complicated way of saying that the curvature tensor relies on the behaviour of second covariant derivatives of a vector field. We might wonder if there's another way of framing this second derivative, using the simple connection symbol grad\boldsymbol{\nabla}. The key is to exploit the similarity between grad\boldsymbol{\nabla} and the exterior derivative d d d\boldsymbol{d}d in their action on vectors.
Originally, we didn't know how to use d d d\boldsymbol{d}d on anything other than forms. In Chapter 34, we defined the connection grad\boldsymbol{\nabla} and the covariant derivative v = v v = v grad_(v)=v*grad\boldsymbol{\nabla}_{\boldsymbol{v}}=\boldsymbol{v} \cdot \boldsymbol{\nabla}v=v. Strip the vector field v v v\boldsymbol{v}v from grad\boldsymbol{\nabla} and its action on a scalar function f f grad f\boldsymbol{\nabla} ff is defined to be equivalent to that of the exterior derivative d f d f df\boldsymbol{d} fdf. We then asked how to use d d d\boldsymbol{d}d on vector fields and came up with the answer that, for vectors, d d d-=grad\boldsymbol{d} \equiv \boldsymbol{\nabla}d. We defined the resulting object d v d v dv\boldsymbol{d} \boldsymbol{v}dv to
be a vector-valued 1-form. The exterior derivative d d d\boldsymbol{d}d acts on the vector field v = v μ e μ v = v μ e μ v=v^(mu)e_(mu)\boldsymbol{v}=v^{\mu} \boldsymbol{e}_{\mu}v=vμeμ one part at a time
(36.4) d v = ( d v μ ) e μ + v μ d e μ (36.4) d v = d v μ e μ + v μ d e μ {:(36.4)dv=(dv^(mu))oxe_(mu)+v^(mu)de_(mu):}\begin{equation*} \boldsymbol{d} \boldsymbol{v}=\left(\boldsymbol{d} v^{\mu}\right) \otimes \boldsymbol{e}_{\mu}+v^{\mu} \boldsymbol{d} \boldsymbol{e}_{\mu} \tag{36.4} \end{equation*}(36.4)dv=(dvμ)eμ+vμdeμ
Let's now slightly diverge from our path in Chapter 34, which next involved writing connection coefficients. Instead, knowing that we are working on a manifold with a connection, we expand the vector-valued 1 -form d e μ d e μ de_(mu)d e_{\mu}deμ, which tells us how the basis vectors change in space. We write this in the (slightly complicated-looking) manner
(36.5) d e μ = ω μ ν e ν (36.5) d e μ = ω μ ν e ν {:(36.5)de_(mu)=omega_(mu)^(nu)oxe_(nu):}\begin{equation*} d e_{\mu}=\omega_{\mu}^{\nu} \otimes e_{\nu} \tag{36.5} \end{equation*}(36.5)deμ=ωμνeν
The objects ω ν μ ω ν μ omega^(nu)_(mu)\omega^{\nu}{ }_{\mu}ωνμ are 1-forms that express the change of the basis vectors in space: they therefore express the connection itself, and we call them connection 1-forms. The right-hand side of eqn 36.5 tells us that, at a particular point in space, the way in which the basis vectors change can be given by an expansion of tensor products of the basis vectors and connection 1-forms ω ν μ ω ν μ omega^(nu)_(mu)\boldsymbol{\omega}^{\nu}{ }_{\mu}ωνμ.
We can relate the connection 1 -forms to connection coefficients, as they describe the same property of the space. We write
(36.6) Γ α μ ν e ν = α e μ = e μ , e α d e μ , e α = ( ω μ ν e ν ) , e α (36.6) Γ α μ ν e ν = α e μ = e μ , e α d e μ , e α = ω μ ν e ν , e α {:(36.6)Gamma_(alpha mu)^(nu)e_(nu)=grad_(alpha)e_(mu)=(:grade_(mu),e_(alpha):)-=(:de_(mu),e_(alpha):)=(:(omega_(mu)^(nu)oxe_(nu)),e_(alpha):):}\begin{equation*} \Gamma_{\alpha \mu}^{\nu} \boldsymbol{e}_{\nu}=\nabla_{\alpha} \boldsymbol{e}_{\mu}=\left\langle\nabla \boldsymbol{e}_{\mu}, \boldsymbol{e}_{\alpha}\right\rangle \equiv\left\langle\boldsymbol{d} e_{\mu}, \boldsymbol{e}_{\alpha}\right\rangle=\left\langle\left(\boldsymbol{\omega}_{\mu}^{\nu} \otimes \boldsymbol{e}_{\nu}\right), \boldsymbol{e}_{\alpha}\right\rangle \tag{36.6} \end{equation*}(36.6)Γαμνeν=αeμ=eμ,eαdeμ,eα=(ωμνeν),eα
We spot that, in order for this to be consistent, the connection 1-forms can be written in terms of connection coefficients as follows:
(36.7) ω μ ν = Γ ν λ μ ω λ . (36.7) ω μ ν = Γ ν λ μ ω λ . {:(36.7)omega_(mu)^(nu)=Gamma^(nu)_(lambda mu)omega^(lambda).:}\begin{equation*} \boldsymbol{\omega}_{\mu}^{\nu}=\Gamma^{\nu}{ }_{\lambda \mu} \boldsymbol{\omega}^{\lambda} . \tag{36.7} \end{equation*}(36.7)ωμν=Γνλμωλ.
We check this in the next example.
Example 36.1
Checking this previous equation makes sense, we have
( ω ν μ e ν ) , e α = Γ ν λ μ ( ω λ e ν ) , e α (from eqn 36.7) = Γ ν λ μ ω λ ( e α ) e ν (inserting e α in the first slot) = Γ ν α μ e ν (using ω λ ( e α ) = δ λ α ) . ω ν μ e ν , e α = Γ ν λ μ ω λ e ν , e α  (from eqn 36.7)  = Γ ν λ μ ω λ e α e ν  (inserting  e α  in the first slot)  = Γ ν α μ e ν  (using  ω λ e α = δ λ α . {:[(:(omega^(nu)_(mu)oxe_(nu)),e_(alpha):)=Gamma^(nu)_(lambda mu)(:(omega^(lambda)oxe_(nu)),e_(alpha):)" (from eqn 36.7) "],[=Gamma^(nu)_(lambda mu)omega^(lambda)(e_(alpha))oxe_(nu)" (inserting "e_(alpha)" in the first slot) "],[=Gamma^(nu)_(alpha mu)e_(nu)" (using "{:omega^(lambda)(e_(alpha))=delta^(lambda)_(alpha)).]:}\begin{aligned} \left\langle\left(\boldsymbol{\omega}^{\nu}{ }_{\mu} \otimes \boldsymbol{e}_{\nu}\right), \boldsymbol{e}_{\alpha}\right\rangle & =\Gamma^{\nu}{ }_{\lambda \mu}\left\langle\left(\boldsymbol{\omega}^{\lambda} \otimes \boldsymbol{e}_{\nu}\right), \boldsymbol{e}_{\alpha}\right\rangle & & \text { (from eqn 36.7) } \\ & =\Gamma^{\nu}{ }_{\lambda \mu} \boldsymbol{\omega}^{\lambda}\left(\boldsymbol{e}_{\alpha}\right) \otimes \boldsymbol{e}_{\nu} & & \text { (inserting } \boldsymbol{e}_{\alpha} \text { in the first slot) } \\ & =\Gamma^{\nu}{ }_{\alpha \mu} \boldsymbol{e}_{\nu} & & \text { (using } \left.\boldsymbol{\omega}^{\lambda}\left(\boldsymbol{e}_{\alpha}\right)=\delta^{\lambda}{ }_{\alpha}\right) . \end{aligned}(ωνμeν),eα=Γνλμ(ωλeν),eα (from eqn 36.7) =Γνλμωλ(eα)eν (inserting eα in the first slot) =Γναμeν (using ωλ(eα)=δλα).
This is exactly what we started with, validating the definitions above.
We now have a 1 -form ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν that expresses the connection. 3 3 ^(3){ }^{3}3 Next, we need to determine the curvature.
Roughly speaking we defined curvature in the last chapter in terms of a double derivative of a field v v v\boldsymbol{v}v by using the operator grad\boldsymbol{\nabla}. Here we shall take the double exterior derivative of v v v\boldsymbol{v}v using d d d\boldsymbol{d}d. The first derivative results in a vector-valued 1 -form, and is given by
(36.8) d v = e μ ( d v μ + ω ν μ v ν ) (36.8) d v = e μ d v μ + ω ν μ v ν {:(36.8)dv=e_(mu)ox(dv^(mu)+omega_(nu)^(mu)v^(nu)):}\begin{equation*} \boldsymbol{d} \boldsymbol{v}=\boldsymbol{e}_{\mu} \otimes\left(\boldsymbol{d} v^{\mu}+\boldsymbol{\omega}_{\nu}^{\mu} v^{\nu}\right) \tag{36.8} \end{equation*}(36.8)dv=eμ(dvμ+ωνμvν)
3 3 ^(3){ }^{3}3 In fact, eqn 36.6 provides us with an interpretation of ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν. The object α e μ α e μ grad_(alpha)e_(mu)\boldsymbol{\nabla}_{\alpha} \boldsymbol{e}_{\mu}αeμ tells us the rate of change of the basis vector e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ along the α α alpha\alphaα direction, and, by the last step in eqn 36.6, that this is equal to ω ν μ ( e α ) e ν ω ν μ e α e ν omega^(nu)_(mu)(e_(alpha))oxe_(nu)\boldsymbol{\omega}^{\nu}{ }_{\mu}\left(\boldsymbol{e}_{\alpha}\right) \otimes \boldsymbol{e}_{\nu}ωνμ(eα)eν. We can therefore interpret ω ν μ ( e α ) ω ν μ e α omega^(nu)_(mu)(e_(alpha))\boldsymbol{\omega}^{\nu}{ }_{\mu}\left(\boldsymbol{e}_{\alpha}\right)ωνμ(eα) as encoding the rate at which e μ μ e μ μ e_(mu)^(mu)\boldsymbol{e}_{\mu}^{\mu}eμμ rotates towards e ν e ν e_(nu)e_{\nu}eν as we move along a curve whose tangent vector is given by e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα.
4 4 ^(4){ }^{4}4 It's important to note that, because the derivative of a vector results in a ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) object built from the tensor product of a vector and a 1 -form (and not the antisymmetric wedge product between two 1 -forms), they generally have non-zero exterior derivatives. In short, vectors allow the existence of the peculiar-looking double exterior derivative and it is this that gives us acces to the curvature.
5 5 ^(5){ }^{5}5 We interpret this as
R ( v ) = ω ν ( v ) e μ R μ ν = ω ν ( e α ) e μ R μ ν v α = ( e μ R μ α ) v α , R ( v ) = ω ν ( v ) e μ R μ ν = ω ν e α e μ R μ ν v α = e μ R μ α v α , {:[R(v)=omega^(nu)(v)oxe_(mu)oxR^(mu)_(nu)],[=omega^(nu)(e_(alpha))e_(mu)oxR^(mu)_(nu)v^(alpha)],[=(e_(mu)oxR^(mu)_(alpha))v^(alpha)","]:}\begin{aligned} \mathcal{R}(\boldsymbol{v}) & =\boldsymbol{\omega}^{\nu}(\boldsymbol{v}) \otimes \boldsymbol{e}_{\mu} \otimes \mathcal{R}^{\mu}{ }_{\nu} \\ & =\boldsymbol{\omega}^{\nu}\left(\boldsymbol{e}_{\alpha}\right) \boldsymbol{e}_{\mu} \otimes \mathcal{R}^{\mu}{ }_{\nu} v^{\alpha} \\ & =\left(\boldsymbol{e}_{\mu} \otimes \mathcal{R}^{\mu}{ }_{\alpha}\right) v^{\alpha}, \end{aligned}R(v)=ων(v)eμRμν=ων(eα)eμRμνvα=(eμRμα)vα,
which is what we had in eqn 36.11.
Differentiate again 4 4 ^(4){ }^{4}4 and we obtain a vector-valued 2-form, given by
d 2 v = d e μ ( d v μ + ω μ ν v ν ) + e μ ( d 2 v μ + d ω μ ν ν ν ω μ ν d v ν ) = e μ ( ω μ ν d v ν + ω μ α ω α ν v ν + d 2 v μ + d ω μ ν v ν (36.9) ω μ ν d v ν ) . d 2 v = d e μ d v μ + ω μ ν v ν + e μ d 2 v μ + d ω μ ν ν ν ω μ ν d v ν = e μ ω μ ν d v ν + ω μ α ω α ν v ν + d 2 v μ + d ω μ ν v ν (36.9) ω μ ν d v ν . {:[d^(2)v=de_(mu)^^(dv^(mu)+omega^(mu)_(nu)v^(nu))+e_(mu)ox(d^(2)v^(mu)+domega^(mu)_(nu)nu^(nu)-omega^(mu)_(nu)^^dv^(nu))],[=e_(mu)ox(omega^(mu)_(nu)^^dv^(nu)+omega^(mu)_(alpha)^^omega^(alpha)_(nu)v^(nu)+d^(2)v^(mu)+domega^(mu)_(nu)v^(nu):}],[(36.9){:-omega^(mu)_(nu)^^dv^(nu)).]:}\begin{align*} \boldsymbol{d}^{2} \boldsymbol{v}= & \boldsymbol{d} \boldsymbol{e}_{\mu} \wedge\left(\boldsymbol{d} v^{\mu}+\boldsymbol{\omega}^{\mu}{ }_{\nu} v^{\nu}\right)+\boldsymbol{e}_{\mu} \otimes\left(\boldsymbol{d}^{2} v^{\mu}+\boldsymbol{d} \boldsymbol{\omega}^{\mu}{ }_{\nu} \nu^{\nu}-\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{d} v^{\nu}\right) \\ = & \boldsymbol{e}_{\mu} \otimes\left(\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{d} v^{\nu}+\boldsymbol{\omega}^{\mu}{ }_{\alpha} \wedge \boldsymbol{\omega}^{\alpha}{ }_{\nu} v^{\nu}+\boldsymbol{d}^{2} v^{\mu}+\boldsymbol{d} \boldsymbol{\omega}^{\mu}{ }_{\nu} v^{\nu}\right. \\ & \left.-\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{d} v^{\nu}\right) . \tag{36.9} \end{align*}d2v=deμ(dvμ+ωμνvν)+eμ(d2vμ+dωμνννωμνdvν)=eμ(ωμνdvν+ωμαωανvν+d2vμ+dωμνvν(36.9)ωμνdvν).
Noting that since the component v μ v μ v^(mu)v^{\mu}vμ is itself merely a scalar function and so d 2 v μ = 0 d 2 v μ = 0 d^(2)v^(mu)=0\boldsymbol{d}^{2} v^{\mu}=0d2vμ=0 (for the usual reason that d 2 = 0 d 2 = 0 d^(2)=0\boldsymbol{d}^{2}=0d2=0 when acting on any function), we conclude that
(36.10) d 2 v = e μ ( d ω μ ν + ω μ α ω α ν ) v ν , (36.10) d 2 v = e μ d ω μ ν + ω μ α ω α ν v ν , {:(36.10)d^(2)v=e_(mu)ox(domega^(mu)_(nu)+omega^(mu)_(alpha)^^omega^(alpha)_(nu))v^(nu)",":}\begin{equation*} \boldsymbol{d}^{2} \boldsymbol{v}=\boldsymbol{e}_{\mu} \otimes\left(\boldsymbol{d} \boldsymbol{\omega}^{\mu}{ }_{\nu}+\boldsymbol{\omega}^{\mu}{ }_{\alpha} \wedge \boldsymbol{\omega}^{\alpha}{ }_{\nu}\right) v^{\nu}, \tag{36.10} \end{equation*}(36.10)d2v=eμ(dωμν+ωμαωαν)vν,
or
(36.11) d 2 v = ( e μ R ν μ ) v ν (36.11) d 2 v = e μ R ν μ v ν {:(36.11)d^(2)v=(e_(mu)oxR_(nu)^(mu))v^(nu):}\begin{equation*} \boldsymbol{d}^{2} \boldsymbol{v}=\left(\boldsymbol{e}_{\mu} \otimes \mathcal{R}_{\nu}^{\mu}\right) v^{\nu} \tag{36.11} \end{equation*}(36.11)d2v=(eμRνμ)vν
where we define the curvature 2 -form R μ ν R μ ν R^(mu)_(nu)\mathcal{R}^{\mu}{ }_{\nu}Rμν by
(36.12) R ν μ = d ω ν μ + ω α μ ω ν α . (36.12) R ν μ = d ω ν μ + ω α μ ω ν α . {:(36.12)R_(nu)^(mu)=domega_(nu)^(mu)+omega_(alpha)^(mu)^^omega_(nu)^(alpha).:}\begin{equation*} \mathcal{R}_{\nu}^{\mu}=\boldsymbol{d} \boldsymbol{\omega}_{\nu}^{\mu}+\boldsymbol{\omega}_{\alpha}^{\mu} \wedge \boldsymbol{\omega}_{\nu}^{\alpha} . \tag{36.12} \end{equation*}(36.12)Rνμ=dωνμ+ωαμωνα.
Equation 36.12 is the most important one of this chapter. It is built from two terms, the first is the exterior derivative of the connection 1forms, expressing how their rate of change in space; the second is a wedge product of the connection 1-forms, expressing their tube-like structure in space. Both of these aspects are needed to describe curvature. We also write, for economy, a curvature 2 -form operator R ( ) R ( ) R()\mathcal{R}()R(), which is a ( 2 , 2 ) ( 2 , 2 ) (2,2)(2,2)(2,2) object designed for the input of a vector 5 5 ^(5){ }^{5}5
(36.13) R ( ) = ω ν ( ) e μ R μ ν . (36.13) R ( ) = ω ν ( ) e μ R μ ν . {:(36.13)R()=omega^(nu)()oxe_(mu)oxR^(mu)_(nu).:}\begin{equation*} \mathcal{R}()=\omega^{\nu}() \otimes e_{\mu} \otimes \mathcal{R}^{\mu}{ }_{\nu} . \tag{36.13} \end{equation*}(36.13)R()=ων()eμRμν.
This allows us to write the memorable equation for the second exterior derivative of a vector in terms of the curvature 2 -form operator as
(36.14) d 2 v = R ( v ) (36.14) d 2 v = R ( v ) {:(36.14)d^(2)v=R(v):}\begin{equation*} d^{2} \boldsymbol{v}=\mathcal{R}(\boldsymbol{v}) \tag{36.14} \end{equation*}(36.14)d2v=R(v)

Example 36.2

Let's prove this is equivalent to the previously defined curvature tensor by contracting the 2 -form R μ ν = d ω μ + ω μ α ω α ν R μ ν = d ω μ + ω μ α ω α ν R^(mu)_(nu)=domega^(mu)+omega^(mu)_(alpha)^^omega^(alpha)_(nu)\mathcal{R}^{\mu}{ }_{\nu}=\boldsymbol{d} \boldsymbol{\omega}^{\mu}+\boldsymbol{\omega}^{\mu}{ }_{\alpha} \wedge \boldsymbol{\omega}^{\alpha}{ }_{\nu}Rμν=dωμ+ωμαωαν with two vectors. First, consider the 1-form
(36.15) ω ν μ = Γ σ ν μ ω σ . (36.15) ω ν μ = Γ σ ν μ ω σ . {:(36.15)omega_(nu)^(mu)=Gamma_(sigma nu)^(mu)omega^(sigma).:}\begin{equation*} \boldsymbol{\omega}_{\nu}^{\mu}=\Gamma_{\sigma \nu}^{\mu} \boldsymbol{\omega}^{\sigma} . \tag{36.15} \end{equation*}(36.15)ωνμ=Γσνμωσ.
Take its exterior derivative
(36.16) d ω ν μ = Γ σ ν μ x λ ω λ ω σ . (36.16) d ω ν μ = Γ σ ν μ x λ ω λ ω σ . {:(36.16)domega_(nu)^(mu)=(delGamma_(sigma nu)^(mu))/(delx^(lambda))omega^(lambda)^^omega^(sigma).:}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}_{\nu}^{\mu}=\frac{\partial \Gamma_{\sigma \nu}^{\mu}}{\partial x^{\lambda}} \boldsymbol{\omega}^{\lambda} \wedge \boldsymbol{\omega}^{\sigma} . \tag{36.16} \end{equation*}(36.16)dωνμ=Γσνμxλωλωσ.
Contract this with vectors e α e α e_(alpha)\boldsymbol{e}_{\alpha}eα and e β e β e_(beta)\boldsymbol{e}_{\beta}eβ as follows:
d ω ν μ ( e α , e β ) = Γ μ σ ν x λ ω λ ω σ ( e α , e β ) = 1 2 Γ μ σ ν x λ [ ω λ ω σ ( e α , e β ) ω σ ω λ ( e α , e β ) ] = 1 2 Γ μ σ ν x λ ( δ α λ δ β σ δ α σ δ β λ ) (36.17) = 1 2 Γ μ β ν x α Γ μ α ν x β . d ω ν μ e α , e β = Γ μ σ ν x λ ω λ ω σ e α , e β = 1 2 Γ μ σ ν x λ ω λ ω σ e α , e β ω σ ω λ e α , e β = 1 2 Γ μ σ ν x λ δ α λ δ β σ δ α σ δ β λ (36.17) = 1 2 Γ μ β ν x α Γ μ α ν x β . {:[domega_(nu)^(mu)(e_(alpha),e_(beta))=(delGamma^(mu)_(sigma nu))/(delx^(lambda))omega^(lambda)^^omega^(sigma)(e_(alpha),e_(beta))],[=(1)/(2)(delGamma^(mu)_(sigma nu))/(delx^(lambda))[omega^(lambda)oxomega^(sigma)(e_(alpha),e_(beta))-omega^(sigma)oxomega^(lambda)(e_(alpha),e_(beta))]],[=(1)/(2)(delGamma^(mu)_(sigma nu))/(delx^(lambda))*(delta_(alpha)^(lambda)delta_(beta)^(sigma)-delta_(alpha)^(sigma)delta_(beta)^(lambda))],[(36.17)=(1)/(2)(delGamma^(mu)_(beta nu))/(delx^(alpha))-(delGamma^(mu)_(alpha nu))/(delx^(beta)).]:}\begin{align*} \boldsymbol{d} \boldsymbol{\omega}_{\nu}^{\mu}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right) & =\frac{\partial \Gamma^{\mu}{ }_{\sigma \nu}}{\partial x^{\lambda}} \boldsymbol{\omega}^{\lambda} \wedge \boldsymbol{\omega}^{\sigma}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right) \\ & =\frac{1}{2} \frac{\partial \Gamma^{\mu}{ }_{\sigma \nu}}{\partial x^{\lambda}}\left[\boldsymbol{\omega}^{\lambda} \otimes \boldsymbol{\omega}^{\sigma}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right)-\boldsymbol{\omega}^{\sigma} \otimes \boldsymbol{\omega}^{\lambda}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right)\right] \\ & =\frac{1}{2} \frac{\partial \Gamma^{\mu}{ }_{\sigma \nu}}{\partial x^{\lambda}} \cdot\left(\delta_{\alpha}^{\lambda} \delta_{\beta}^{\sigma}-\delta_{\alpha}^{\sigma} \delta_{\beta}^{\lambda}\right) \\ & =\frac{1}{2} \frac{\partial \Gamma^{\mu}{ }_{\beta \nu}}{\partial x^{\alpha}}-\frac{\partial \Gamma^{\mu}{ }_{\alpha \nu}}{\partial x^{\beta}} . \tag{36.17} \end{align*}dωνμ(eα,eβ)=Γμσνxλωλωσ(eα,eβ)=12Γμσνxλ[ωλωσ(eα,eβ)ωσωλ(eα,eβ)]=12Γμσνxλ(δαλδβσδασδβλ)(36.17)=12ΓμβνxαΓμανxβ.
Now contract the second term ω μ σ ω σ ν ω μ σ ω σ ν omega^(mu)_(sigma)^^omega^(sigma)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\sigma} \wedge \boldsymbol{\omega}^{\sigma}{ }_{\nu}ωμσωσν in the same way
ω σ μ ω ν σ ( e α , e β ) = 1 2 [ ( ω σ μ ω σ ν ω ν σ ω μ σ ) ( e α , e β ) ] = 1 2 [ Γ μ λ σ Γ σ ρ ν ω λ ( e α ) ω ρ ( e β ) Γ σ ρ ν Γ μ λ σ ω ρ ( e α ) ω λ ( e β ) ] (36.18) = ( Γ μ α σ Γ σ β ν Γ σ α ν Γ μ β σ ) . ω σ μ ω ν σ e α , e β = 1 2 ω σ μ ω σ ν ω ν σ ω μ σ e α , e β = 1 2 Γ μ λ σ Γ σ ρ ν ω λ e α ω ρ e β Γ σ ρ ν Γ μ λ σ ω ρ e α ω λ e β (36.18) = Γ μ α σ Γ σ β ν Γ σ α ν Γ μ β σ . {:[omega_(sigma)^(mu)^^omega_(nu)^(sigma)(e_(alpha),e_(beta))=(1)/(2)[(omega_(sigma)^(mu)ox_(omega^(sigma))_(nu)-omega_(nu)^(sigma)oxomega^(mu)_(sigma))(e_(alpha),e_(beta))]],[=(1)/(2)[Gamma^(mu)_(lambda sigma)Gamma^(sigma)_(rho nu)omega^(lambda)(e_(alpha))oxomega^(rho)(e_(beta))-Gamma^(sigma)_(rho nu)Gamma^(mu)_(lambda sigma)omega^(rho)(e_(alpha))oxomega^(lambda)(e_(beta))]],[(36.18)=(Gamma^(mu)_(alpha sigma)Gamma^(sigma)_(beta nu)-Gamma^(sigma)_(alpha nu)Gamma^(mu)_(beta sigma)).]:}\begin{align*} \boldsymbol{\omega}_{\sigma}^{\mu} \wedge \boldsymbol{\omega}_{\nu}^{\sigma}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right) & =\frac{1}{2}\left[\left(\boldsymbol{\omega}_{\sigma}^{\mu} \otimes_{\boldsymbol{\omega}^{\sigma}}{ }_{\nu}-\boldsymbol{\omega}_{\nu}^{\sigma} \otimes \boldsymbol{\omega}^{\mu}{ }_{\sigma}\right)\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right)\right] \\ & =\frac{1}{2}\left[\Gamma^{\mu}{ }_{\lambda \sigma} \Gamma^{\sigma}{ }_{\rho \nu} \boldsymbol{\omega}^{\lambda}\left(\boldsymbol{e}_{\alpha}\right) \otimes \boldsymbol{\omega}^{\rho}\left(\boldsymbol{e}_{\beta}\right)-\Gamma^{\sigma}{ }_{\rho \nu} \Gamma^{\mu}{ }_{\lambda \sigma} \boldsymbol{\omega}^{\rho}\left(\boldsymbol{e}_{\alpha}\right) \otimes \boldsymbol{\omega}^{\lambda}\left(\boldsymbol{e}_{\beta}\right)\right] \\ & =\left(\Gamma^{\mu}{ }_{\alpha \sigma} \Gamma^{\sigma}{ }_{\beta \nu}-\Gamma^{\sigma}{ }_{\alpha \nu} \Gamma^{\mu}{ }_{\beta \sigma}\right) . \tag{36.18} \end{align*}ωσμωνσ(eα,eβ)=12[(ωσμωσνωνσωμσ)(eα,eβ)]=12[ΓμλσΓσρνωλ(eα)ωρ(eβ)ΓσρνΓμλσωρ(eα)ωλ(eβ)](36.18)=(ΓμασΓσβνΓσανΓμβσ).
We conclude that
R ν μ ( e α , e β ) = 1 2 ( Γ μ β ν x α Γ μ α ν x β + Γ μ α σ Γ σ β ν Γ β σ μ Γ α ν σ ) (36.19) = 1 2 R ν α β μ . R ν μ e α , e β = 1 2 Γ μ β ν x α Γ μ α ν x β + Γ μ α σ Γ σ β ν Γ β σ μ Γ α ν σ (36.19) = 1 2 R ν α β μ . {:[R_(nu)^(mu)(e_(alpha),e_(beta))=(1)/(2)((delGamma^(mu)_(beta nu))/(delx^(alpha))-(delGamma^(mu)_(alpha nu))/(delx^(beta))+Gamma^(mu)_(alpha sigma)Gamma^(sigma)_(beta nu)-Gamma_(beta sigma)^(mu)Gamma_(alpha nu)^(sigma))],[(36.19)=(1)/(2)R_(nu alpha beta)^(mu).]:}\begin{align*} \mathcal{R}_{\nu}^{\mu}\left(\boldsymbol{e}_{\alpha}, \boldsymbol{e}_{\beta}\right) & =\frac{1}{2}\left(\frac{\partial \Gamma^{\mu}{ }_{\beta \nu}}{\partial x^{\alpha}}-\frac{\partial \Gamma^{\mu}{ }_{\alpha \nu}}{\partial x^{\beta}}+\Gamma^{\mu}{ }_{\alpha \sigma} \Gamma^{\sigma}{ }_{\beta \nu}-\Gamma_{\beta \sigma}^{\mu} \Gamma_{\alpha \nu}^{\sigma}\right) \\ & =\frac{1}{2} R_{\nu \alpha \beta}^{\mu} . \tag{36.19} \end{align*}Rνμ(eα,eβ)=12(ΓμβνxαΓμανxβ+ΓμασΓσβνΓβσμΓανσ)(36.19)=12Rναβμ.
From the last example we see that we can write the curvature 2 -form R μ ν R μ ν R^(mu)_(nu)\mathcal{R}^{\mu}{ }_{\nu}Rμν in terms of the components of the Riemann tensor R R R\boldsymbol{R}R :
(36.20) R ν μ = R ν | α β | μ ω α ω β , (36.20) R ν μ = R ν | α β | μ ω α ω β , {:(36.20)R_(nu)^(mu)=R_(nu|alpha beta|)^(mu)omega^(alpha)^^omega^(beta)",":}\begin{equation*} \mathcal{R}_{\nu}^{\mu}=R_{\nu|\alpha \beta|}^{\mu} \boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta}, \tag{36.20} \end{equation*}(36.20)Rνμ=Rν|αβ|μωαωβ,
where the restriction on the sum removes the factor of 1 / 2 1 / 2 1//21 / 21/2 that we had above. 6 6 ^(6){ }^{6}6 We now have an expression for the curvature 2 -form, but lack a simple method for finding the connection 1-forms ω μ α ω μ α omega^(mu)_(alpha)\boldsymbol{\omega}^{\mu}{ }_{\alpha}ωμα from the metric field g g g\boldsymbol{g}g. We now turn to this matter, since it's our reason for pursuing this formalism.

36.2 Two rules

The previous section resulted in a number of equivalent expressions designed to compute the curvature of a spacetime. They rely on using the connection 1 -forms of the spacetime. So how do we find the connection 1 -forms ω μ ν ω μ ν omega^(mu)_(nu)\boldsymbol{\omega}^{\mu}{ }_{\nu}ωμν ? The good news is that, given the metric, they can usually be found by guesswork! More specifically, we express our ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) metric field g g g\boldsymbol{g}g in terms of basis 1-forms g = g μ ν ω μ ω ν g = g μ ν ω μ ω ν g=g_(mu nu)omega^(mu)oxomega^(nu)\boldsymbol{g}=g_{\mu \nu} \boldsymbol{\omega}^{\mu} \otimes \boldsymbol{\omega}^{\nu}g=gμνωμων and use the exterior derivatives of the basis 1 -forms to guess the connection 1 -forms.
What enables this strategy are two very restrictive rules that we can derive by simply considering the action of a Leibniz product rule on a dot product
(36.22) d ( u v ) = ( d u ) v + u ( d v ) (36.22) d ( u v ) = ( d u ) v + u ( d v ) {:(36.22)d(u*v)=(du)*v+u*(dv):}\begin{equation*} \boldsymbol{d}(\boldsymbol{u} \cdot \boldsymbol{v})=(\boldsymbol{d} \boldsymbol{u}) \cdot \boldsymbol{v}+\boldsymbol{u} \cdot(\boldsymbol{d} \boldsymbol{v}) \tag{36.22} \end{equation*}(36.22)d(uv)=(du)v+u(dv)
We're familiar with the important rule ω μ , e ν = δ μ ν ω μ , e ν = δ μ ν (:omega^(mu),e_(nu):)=delta^(mu)_(nu)\left\langle\boldsymbol{\omega}^{\mu}, \boldsymbol{e}_{\nu}\right\rangle=\delta^{\mu}{ }_{\nu}ωμ,eν=δμν. The conceptual trick here, due to Cartan, is to take a field of vectors and a field of 1 -forms and consider the contraction of a basis vector e μ e μ e_(mu)\boldsymbol{e}_{\mu}eμ and a basis 1 -form ω μ ω μ omega^(mu)\boldsymbol{\omega}^{\mu}ωμ, at a specific point P P P\mathcal{P}P. In this spirit, we write that the derivative of a point d P d P dP\boldsymbol{d} \mathcal{P}dP corresponds to
(36.23) d P = e μ ω μ . (36.23) d P = e μ ω μ . {:(36.23)dP=e_(mu)oxomega^(mu).:}\begin{equation*} \boldsymbol{d} \mathcal{P}=e_{\mu} \otimes \boldsymbol{\omega}^{\mu} . \tag{36.23} \end{equation*}(36.23)dP=eμωμ.
What does this mean and how does this work? Cartan's view was that a vector could be seen as the movement of a point and so the most primitive version of a vector would be the movement of a point d P d P dP\boldsymbol{d} \mathcal{P}dP.
6 6 ^(6){ }^{6}6 The | α β | | α β | |alpha beta||\alpha \beta||αβ| notation here should be interpreted as only allowing components in the order α β α β alpha beta\alpha \betaαβ and not β α β α beta alpha\beta \alphaβα. In the interests of writing memorable equations, we can also write things in terms of the R ( R ( R(\mathcal{R}(R( ) operator, but defining the ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) R μ ν R μ ν R^(mu nu)\mathcal{R}^{\mu \nu}Rμν, such that R ( ) = 1 2 ( e μ e ν ) R μ ν R ( ) = 1 2 e μ e ν R μ ν R()=(1)/(2)(e_(mu)^^e_(nu))R^(mu nu)\mathcal{R}()=\frac{1}{2}\left(\boldsymbol{e}_{\mu} \wedge \boldsymbol{e}_{\nu}\right) \mathcal{R}^{\mu \nu}R()=12(eμeν)Rμν. R R R\mathcal{R}R, such that R ( ) = 1 2 ( e μ e ν ) R R μ ν R ( ) = 1 2 e μ e ν R R μ ν R()=(1)/(2)(e_(mu)^^e_(nu))RR^(mu nu)\mathcal{R}()=\frac{1}{2}\left(\boldsymbol{e}_{\mu} \wedge \boldsymbol{e}_{\nu}\right) \mathcal{R} \mathcal{R}^{\mu \nu}R()=12(eμeν)RRμν.
Then we can write the curvature operThen we can write the curvature oper-
ator in terms of the components of the ator in terms of the components of the
Riemann tensor. The curvature operator R R R\mathcal{R}R is given by
R ( ) = 1 4 ( e μ e ν ) R μ ν α β ( ω α ω β ) R ( ) = 1 4 e μ e ν R μ ν α β ω α ω β R()=(1)/(4)(e_(mu)^^e_(nu))R^(mu nu)_(alpha beta)(omega^(alpha)^^omega^(beta))\mathcal{R}()=\frac{1}{4}\left(\boldsymbol{e}_{\mu} \wedge \boldsymbol{e}_{\nu}\right) R^{\mu \nu}{ }_{\alpha \beta}\left(\boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta}\right)R()=14(eμeν)Rμναβ(ωαωβ).
7 7 ^(7){ }^{7}7 That is to say
e μ ω μ ( v ) = e μ v μ = v e μ ω μ ( v ) = e μ v μ = v {:[e_(mu)oxomega^(mu)(v)=e_(mu)v^(mu)],[=v]:}\begin{aligned} \boldsymbol{e}_{\mu} \otimes \boldsymbol{\omega}^{\mu}(\boldsymbol{v}) & =\boldsymbol{e}_{\mu} v^{\mu} \\ & =\boldsymbol{v} \end{aligned}eμωμ(v)=eμvμ=v
This is not yet quite a vector. The object d P ( d P ( dP(\boldsymbol{d} \mathcal{P}(dP(, ) s h o u l d b e i n t e r p r e t e d ) s h o u l d b e i n t e r p r e t e d )shouldbeinterpreted) should be interpreted)shouldbeinterpreted as a ( 1 , 1 ) ( 1 , 1 ) (1,1)(1,1)(1,1) quantity. This means that it's a vector that hasn't yet been given a direction, and so is a sort of proto-vector that doesn't point anywhere. In fact, to point it along a direction we insert a vector v v vvv into one of its slots, to find d P , v = v d P , v = v (:dP,v:)=v\langle\boldsymbol{d} \mathcal{P}, \boldsymbol{v}\rangle=\boldsymbol{v}dP,v=v. For this to work, we must therefore have that 7 d P = e μ ω μ 7 d P = e μ ω μ ^(7)dP=e_(mu)oxomega^(mu){ }^{7} \boldsymbol{d} \mathcal{P}=\boldsymbol{e}_{\mu} \otimes \boldsymbol{\omega}^{\mu}7dP=eμωμ.
Next, we apply the product rule to the exterior derivative of d P = d P = dP=\boldsymbol{d} \mathcal{P}=dP= e μ ω μ e μ ω μ e_(mu)oxomega^(mu)\boldsymbol{e}_{\mu} \otimes \boldsymbol{\omega}^{\mu}eμωμ. Since P P P\mathcal{P}P acts like a function, we must have d 2 P = 0 d 2 P = 0 d^(2)P=0\boldsymbol{d}^{2} \mathcal{P}=0d2P=0, and so
0 = d 2 P = d e μ ω μ + e μ d ω μ (36.25) = e μ ( ω μ ν ω ν + d ω μ ) 0 = d 2 P = d e μ ω μ + e μ d ω μ (36.25) = e μ ω μ ν ω ν + d ω μ {:[0=d^(2)P=de_(mu)^^omega^(mu)+e_(mu)ox domega^(mu)],[(36.25)=e_(mu)ox(omega^(mu)_(nu)^^omega^(nu)+domega^(mu))]:}\begin{align*} 0 & =\boldsymbol{d}^{2} \mathcal{P}=\boldsymbol{d} e_{\mu} \wedge \boldsymbol{\omega}^{\mu}+\boldsymbol{e}_{\mu} \otimes \boldsymbol{d} \boldsymbol{\omega}^{\mu} \\ & =\boldsymbol{e}_{\mu} \otimes\left(\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{\omega}^{\nu}+\boldsymbol{d} \boldsymbol{\omega}^{\mu}\right) \tag{36.25} \end{align*}0=d2P=deμωμ+eμdωμ(36.25)=eμ(ωμνων+dωμ)
8 8 ^(8){ }^{8}8 This is also sometimes known as Cartan's first structural equation. Note that combining it with ω ν μ = Γ ν λ μ ω λ ω ν μ = Γ ν λ μ ω λ omega^(nu)_(mu)=Gamma^(nu)_(lambda mu)omega^(lambda)\boldsymbol{\omega}^{\nu}{ }_{\mu}=\Gamma^{\nu}{ }_{\lambda \mu} \boldsymbol{\omega}^{\lambda}ωνμ=Γνλμωλ we obtain
d ω μ = Γ β ν μ ω ν ω β d ω μ = Γ β ν μ ω ν ω β domega^(mu)=Gamma_(beta nu)^(mu)omega^(nu)^^omega^(beta)\boldsymbol{d} \boldsymbol{\omega}^{\mu}=\Gamma_{\beta \nu}^{\mu} \boldsymbol{\omega}^{\nu} \wedge \boldsymbol{\omega}^{\beta}dωμ=Γβνμωνωβ
as we claimed in Chapter 34. We can compare this with
(36.27) d e μ = Γ β μ ν ω β e ν (36.27) d e μ = Γ β μ ν ω β e ν {:(36.27)de_(mu)=Gamma_(beta mu)^(nu)omega^(beta)oxe_(nu):}\begin{equation*} \boldsymbol{d} \boldsymbol{e}_{\mu}=\Gamma_{\beta \mu}^{\nu} \boldsymbol{\omega}^{\beta} \otimes \boldsymbol{e}_{\nu} \tag{36.27} \end{equation*}(36.27)deμ=Γβμνωβeν
which follows from eqn 36.5, and (also from Chapter 34)
(36.28) ω μ = Γ β ν μ ω ν ω β . (36.28) ω μ = Γ β ν μ ω ν ω β . {:(36.28)gradomega^(mu)=-Gamma_(beta nu)^(mu)omega^(nu)oxomega^(beta).:}\begin{equation*} \boldsymbol{\nabla} \boldsymbol{\omega}^{\mu}=-\Gamma_{\beta \nu}^{\mu} \boldsymbol{\omega}^{\nu} \otimes \boldsymbol{\omega}^{\beta} . \tag{36.28} \end{equation*}(36.28)ωμ=Γβνμωνωβ.
This gives us the first of our two rules: a restrictive equation known as 8 8 ^(8){ }^{8}8 the symmetry condition (rule 1)
(36.29) d ω μ + ω ν μ ω ν = 0 (36.29) d ω μ + ω ν μ ω ν = 0 {:(36.29)domega^(mu)+omega_(nu)^(mu)^^omega^(nu)=0:}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\mu}+\boldsymbol{\omega}_{\nu}^{\mu} \wedge \boldsymbol{\omega}^{\nu}=0 \tag{36.29} \end{equation*}(36.29)dωμ+ωνμων=0
There is second simplification we can make to restrict the connection 1 -forms. This one is based on the definition g μ ν = e μ e ν g μ ν = e μ e ν g_(mu nu)=e_(mu)*e_(nu)g_{\mu \nu}=\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}gμν=eμeν. Insert u = e μ u = e μ u=e_(mu)\boldsymbol{u}=\boldsymbol{e}_{\mu}u=eμ and v = e ν v = e ν v=e_(nu)\boldsymbol{v}=\boldsymbol{e}_{\nu}v=eν into the product rule (eqn 36.22) to find
d g μ ν = d ( e μ e ν ) = d e μ e ν + e μ d e ν (36.30) = ( ω μ σ e σ ) e ν + e μ ( ω ν σ e σ ) d g μ ν = d e μ e ν = d e μ e ν + e μ d e ν (36.30) = ω μ σ e σ e ν + e μ ω ν σ e σ {:[dg_(mu nu)=d(e_(mu)*e_(nu))=de_(mu)*e_(nu)+e_(mu)*de_(nu)],[(36.30)=(omega_(mu)^(sigma)oxe_(sigma))*e_(nu)+e_(mu)*(omega_(nu)^(sigma)oxe_(sigma))]:}\begin{align*} \boldsymbol{d} g_{\mu \nu}=\boldsymbol{d}\left(\boldsymbol{e}_{\mu} \cdot e_{\nu}\right) & =\boldsymbol{d} e_{\mu} \cdot e_{\nu}+e_{\mu} \cdot d e_{\nu} \\ & =\left(\boldsymbol{\omega}_{\mu}^{\sigma} \otimes e_{\sigma}\right) \cdot e_{\nu}+e_{\mu} \cdot\left(\boldsymbol{\omega}_{\nu}^{\sigma} \otimes e_{\sigma}\right) \tag{36.30} \end{align*}dgμν=d(eμeν)=deμeν+eμdeν(36.30)=(ωμσeσ)eν+eμ(ωνσeσ)
Again using the fact that a dot is an instruction to combine vectors according to the rule e μ e ν = g μ ν e μ e ν = g μ ν e_(mu)*e_(nu)=g_(mu nu)\boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}=g_{\mu \nu}eμeν=gμν, we obtain
(36.31) d g μ ν = ω μ σ g σ ν + g μ σ ω ν σ (36.31) d g μ ν = ω μ σ g σ ν + g μ σ ω ν σ {:(36.31)dg_(mu nu)=omega_(mu)^(sigma)g_(sigma nu)+g_(mu sigma)omega_(nu)^(sigma):}\begin{equation*} \boldsymbol{d} g_{\mu \nu}=\boldsymbol{\omega}_{\mu}^{\sigma} g_{\sigma \nu}+g_{\mu \sigma} \boldsymbol{\omega}_{\nu}^{\sigma} \tag{36.31} \end{equation*}(36.31)dgμν=ωμσgσν+gμσωνσ
Finally, we lower indices with the metric components and tidy up to obtain our second rule, the compatibility condition (rule 2)
(36.32) d g μ ν = ω μ ν + ω ν μ (36.32) d g μ ν = ω μ ν + ω ν μ {:(36.32)dg_(mu nu)=omega_(mu nu)+omega_(nu mu):}\begin{equation*} \boldsymbol{d} g_{\mu \nu}=\boldsymbol{\omega}_{\mu \nu}+\boldsymbol{\omega}_{\nu \mu} \tag{36.32} \end{equation*}(36.32)dgμν=ωμν+ωνμ
The compatibility condition is actually equivalent to the condition g = g = grad g=\boldsymbol{\nabla} \boldsymbol{g}=g= 0 seen in Chapter 34.
With these two key equations available to help us turn the metric into the connection 1 -forms, we make a final simplifying move. We agree to work in an orthonormal basis, which means g μ ν = η μ ^ ν ^ g μ ν = η μ ^ ν ^ g_(mu nu)=eta_( hat(mu) hat(nu))g_{\mu \nu}=\eta_{\hat{\mu} \hat{\nu}}gμν=ημ^ν^ and d η μ ^ ν ^ = 0 d η μ ^ ν ^ = 0 deta_( hat(mu) hat(nu))=0\boldsymbol{d} \eta_{\hat{\mu} \hat{\nu}}=0dημ^ν^=0, so, by the compatibility condition, we have connections ω μ ^ ν ^ = ω ν ^ μ ^ ω μ ^ ν ^ = ω ν ^ μ ^ omega_( hat(mu) hat(nu))=-omega_( hat(nu) hat(mu))\boldsymbol{\omega}_{\hat{\mu} \hat{\nu}}=-\boldsymbol{\omega}_{\hat{\nu} \hat{\mu}}ωμ^ν^=ων^μ^.
Having covered a lot of technical ground, let's collect together the key ideas, in the order they will be used:
Idea 1: Cartan's two rules for finding curvature 1-forms
0 = d ω μ + ω μ ν ω ν , (36.33) d η μ ^ ν ^ = 0 = ω μ ^ ν ^ + ω ν ^ μ ^ . 0 = d ω μ + ω μ ν ω ν , (36.33) d η μ ^ ν ^ = 0 = ω μ ^ ν ^ + ω ν ^ μ ^ . {:[0=domega^(mu)+omega^(mu)_(nu)^^omega^(nu)","],[(36.33)deta_( hat(mu) hat(nu))=0=omega_( hat(mu) hat(nu))+omega_( hat(nu) hat(mu)).]:}\begin{align*} 0 & =\boldsymbol{d} \boldsymbol{\omega}^{\mu}+\boldsymbol{\omega}^{\mu}{ }_{\nu} \wedge \boldsymbol{\omega}^{\nu}, \\ \boldsymbol{d} \eta_{\hat{\mu} \hat{\nu}}=0 & =\boldsymbol{\omega}_{\hat{\mu} \hat{\nu}}+\boldsymbol{\omega}_{\hat{\nu} \hat{\mu}} . \tag{36.33} \end{align*}0=dωμ+ωμνων,(36.33)dημ^ν^=0=ωμ^ν^+ων^μ^.
Idea 2: The curvature 2-form in terms of connection 1-forms
(36.34) R ν μ = d ω ν μ + ω μ α ω α ν . (36.34) R ν μ = d ω ν μ + ω μ α ω α ν . {:(36.34)R_(nu)^(mu)=domega_(nu)^(mu)+omega^(mu)_(alpha)^^omega^(alpha)_(nu).:}\begin{equation*} \mathcal{R}_{\nu}^{\mu}=\boldsymbol{d} \boldsymbol{\omega}_{\nu}^{\mu}+{\omega^{\mu}}_{\alpha} \wedge \boldsymbol{\omega}^{\alpha}{ }_{\nu} . \tag{36.34} \end{equation*}(36.34)Rνμ=dωνμ+ωμαωαν.
Idea 3: The components of the Riemann tensor
(36.35) R ν μ = R ν | α β | μ ω α ω β . (36.35) R ν μ = R ν | α β | μ ω α ω β . {:(36.35)R_(nu)^(mu)=R_(nu|alpha beta|)^(mu)omega^(alpha)^^omega^(beta).:}\begin{equation*} \mathcal{R}_{\nu}^{\mu}=R_{\nu|\alpha \beta|}^{\mu} \boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta} . \tag{36.35} \end{equation*}(36.35)Rνμ=Rν|αβ|μωαωβ.
As advertised, we shall find in the examples below that Idea 1 and some (rather uninspired) guesswork solves the problem of finding the connection 1-forms ω μ ^ ν ^ ω μ ^ ν ^ omega^( hat(mu))_( hat(nu))\boldsymbol{\omega}^{\hat{\mu}}{ }_{\hat{\nu}}ωμ^ν^. We then immediately have access to the curvature 2 -form and can extract the components of the Riemann tensor. Although, at this stage, this no doubt looks rather involved, in practice, this makes working out Riemann tensor components a fairly simple job.

36.3 Le repère mobile

Cartan intended his method to be used in an orthonormal frame of reference. Although we have been referring regularly to the orthonormal frame, we should perhaps have spoken of orthonormal frames. The idea, after all, is that, at a particular point in spacetime, we erect a local frame of reference and normalize its orthogonal axes. This gives us an orthonormal frame at each point in spacetime, which we might helpfully think of as a field of frames. The relationship between the orthonormal frames found at each point is encoded in the connection, expressed in Cartan's method as the 1-form field ω μ ν ( x ) ω μ ν ( x ) omega^(mu)_(nu)(x)\boldsymbol{\omega}^{\mu}{ }_{\nu}(x)ωμν(x). Cartan himself imagined moving through space watching the orthonormal frame picked out at each point seemingly change as a function of position. He called this changing frame of reference Le repère mobile ( ~~\approx the moving frame). 9 9 ^(9){ }^{9}9 Whichever way we think of it, the field of orthonormal frames is the starting point for using Cartan's method. 10 10 ^(10){ }^{10}10 Once we have the orthonormal frame(s), we can find the connection 1 -forms.
In what follows, we shall be working in local orthonormal frames for our computations. These are frames in which the Minkowski metric raises and lowers indices, but where the connection coefficients do not vanish. As we've stressed previously, there is a simple prescription for obtaining the orthonormal frame: we choose the one appropriate for an observer at rest in the coordinate frame at a particular point.
In general, the metric can be written as
(36.36) g = g μ ν ω μ ω ν (36.36) g = g μ ν ω μ ω ν {:(36.36)g=g_(mu nu)omega^(mu)oxomega^(nu):}\begin{equation*} \boldsymbol{g}=g_{\mu \nu} \boldsymbol{\omega}^{\mu} \otimes \boldsymbol{\omega}^{\nu} \tag{36.36} \end{equation*}(36.36)g=gμνωμων
9 9 ^(9){ }^{9}9 It is somehow appropriate that the French reflexive verb se repérer means to find one's way around, something we are needing to do all the time in curved spaces!
10 10 ^(10){ }^{10}10 In reviewing this procedure here, we are just upgrading the procedure of normalizing the components of a diagonal malizing the comp of diagonal metric, discussed in Chapter 10.
Fig. 36.1 The stages in Cartan's recipe for computing the Riemann tensor.
Should you want them, you can also extract connection coefficients using ω ν ^ μ ^ = Γ ν ^ ı ^ ω λ ω ν ^ μ ^ = Γ ν ^ ı ^ ω λ omega^( hat(nu))_( hat(mu))=Gamma^( hat(nu))_( hat(ı))omega^(lambda)\boldsymbol{\omega}^{\hat{\nu}}{ }_{\hat{\mu}}=\Gamma^{\hat{\nu}}{ }_{\hat{\imath}} \boldsymbol{\omega}^{\lambda}ων^μ^=Γν^ı^ωλ. However, you should remember that, not only are these not remember that, not only are these no required to access the curvature, bu that they cannot simply be transformed between frames as tensors and that, importantly here, in non-coordinate frames we lose the symmetry in the lower indices.
From here we write ω μ ω μ = ( ω μ ) 2 ω μ ω μ = ω μ 2 omega^(mu)oxomega^(mu)=(omega^(mu))^(2)\boldsymbol{\omega}^{\mu} \otimes \boldsymbol{\omega}^{\mu}=\left(\boldsymbol{\omega}^{\mu}\right)^{2}ωμωμ=(ωμ)2, so we have, for the usual case of a diagonal metric in (3+1)-dimensional spacetime,
(36.37) g = g t t ( d t ) 2 + g r r ( d r ) 2 + g θ θ ( d θ ) 2 + g ϕ ϕ ( d ϕ ) 2 (36.37) g = g t t ( d t ) 2 + g r r ( d r ) 2 + g θ θ ( d θ ) 2 + g ϕ ϕ ( d ϕ ) 2 {:(36.37)g=g_(tt)(dt)^(2)+g_(rr)(dr)^(2)+g_(theta theta)(d theta)^(2)+g_(phi phi)(d phi)^(2):}\begin{equation*} \boldsymbol{g}=g_{t t}(\boldsymbol{d} t)^{2}+g_{r r}(\boldsymbol{d} r)^{2}+g_{\theta \theta}(\boldsymbol{d} \theta)^{2}+g_{\phi \phi}(\boldsymbol{d} \phi)^{2} \tag{36.37} \end{equation*}(36.37)g=gtt(dt)2+grr(dr)2+gθθ(dθ)2+gϕϕ(dϕ)2
where we've arbitrarily chosen to use spherical polars. The orthonormal frame has a metric
(36.38) g = ( ω t ^ ) 2 + ( ω r ^ ) 2 + ( ω θ ^ ) 2 + ( ω ϕ ^ ) 2 (36.38) g = ω t ^ 2 + ω r ^ 2 + ω θ ^ 2 + ω ϕ ^ 2 {:(36.38)g=-(omega^( hat(t)))^(2)+(omega^( hat(r)))^(2)+(omega^( hat(theta)))^(2)+(omega^( hat(phi)))^(2):}\begin{equation*} \boldsymbol{g}=-\left(\boldsymbol{\omega}^{\hat{t}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{r}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\theta}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\phi}}\right)^{2} \tag{36.38} \end{equation*}(36.38)g=(ωt^)2+(ωr^)2+(ωθ^)2+(ωϕ^)2
Immediately, therefore, we must have
(36.39) ω t ^ = g t t d t , ω r ^ = g r r d r , ω θ ^ = g θ θ d θ , ω ϕ ^ = g ϕ ϕ d ϕ . (36.39) ω t ^ = g t t d t , ω r ^ = g r r d r , ω θ ^ = g θ θ d θ , ω ϕ ^ = g ϕ ϕ d ϕ . {:(36.39)omega^( hat(t))=sqrt(-g_(tt))dt","quadomega^( hat(r))=sqrt(g_(rr))dr","quadomega^( hat(theta))=sqrt(g_(theta theta))d theta","quadomega^( hat(phi))=sqrt(g_(phi phi))d phi.:}\begin{equation*} \boldsymbol{\omega}^{\hat{t}}=\sqrt{-g_{t t}} \boldsymbol{d} t, \quad \boldsymbol{\omega}^{\hat{r}}=\sqrt{g_{r r}} \boldsymbol{d} r, \quad \boldsymbol{\omega}^{\hat{\theta}}=\sqrt{g_{\theta \theta}} \boldsymbol{d} \theta, \quad \boldsymbol{\omega}^{\hat{\phi}}=\sqrt{g_{\phi \phi}} \boldsymbol{d} \phi . \tag{36.39} \end{equation*}(36.39)ωt^=gttdt,ωr^=grrdr,ωθ^=gθθdθ,ωϕ^=gϕϕdϕ.
This corresponds to the orthonormal frame with basis vectors
(36.40) e t ^ = 1 g t t t , e r ^ = 1 g r r r , e θ ^ = 1 g θ θ θ , e ϕ ^ = 1 g ϕ ϕ ϕ (36.40) e t ^ = 1 g t t t , e r ^ = 1 g r r r , e θ ^ = 1 g θ θ θ , e ϕ ^ = 1 g ϕ ϕ ϕ {:(36.40)e_( hat(t))=(1)/(sqrt(-g_(tt)))(del)/(del t)","quade_( hat(r))=(1)/(sqrt(g_(rr)))(del)/(del r)","quade_( hat(theta))=(1)/(sqrt(g_(theta theta)))(del)/(del theta)","quade_( hat(phi))=(1)/(sqrt(g_(phi phi)))(del)/(del phi):}\begin{equation*} e_{\hat{t}}=\frac{1}{\sqrt{-g_{t t}}} \frac{\partial}{\partial t}, \quad e_{\hat{r}}=\frac{1}{\sqrt{g_{r r}}} \frac{\partial}{\partial r}, \quad e_{\hat{\theta}}=\frac{1}{\sqrt{g_{\theta \theta}}} \frac{\partial}{\partial \theta}, \quad e_{\hat{\phi}}=\frac{1}{\sqrt{g_{\phi \phi}}} \frac{\partial}{\partial \phi} \tag{36.40} \end{equation*}(36.40)et^=1gttt,er^=1grrr,eθ^=1gθθθ,eϕ^=1gϕϕϕ
These latter two equations allow us to read off the non-zero vielbein components
( e t ) t ^ = g t t , ( e r ) r ^ = g r r , ( e θ ) θ ^ = g θ θ , ( e ϕ ) ϕ ^ = g ϕ ϕ , ( e t ^ ) t = 1 g t t , ( e r ^ ) r = 1 g r r , ( e θ ^ ) θ = 1 g θ θ , ( e ϕ ^ ) ϕ = 1 g ϕ ϕ . e t t ^ = g t t , e r r ^ = g r r , e θ θ ^ = g θ θ , e ϕ ϕ ^ = g ϕ ϕ , e t ^ t = 1 g t t , e r ^ r = 1 g r r , e θ ^ θ = 1 g θ θ , e ϕ ^ ϕ = 1 g ϕ ϕ . {:[(e_(t))^( hat(t))=sqrt(-g_(tt))",",(e_(r))^( hat(r))=sqrt(g_(rr))","quad(e_(theta))^( hat(theta))=sqrt(g_(theta theta))","quad(e_(phi))^( hat(phi))=sqrt(g_(phi phi))","],[(e_( hat(t)))^(t)=(1)/(sqrt(-g_(tt)))","quad(e_( hat(r)))^(r)=(1)/(sqrt(g_(rr)))","quad(e_( hat(theta)))^(theta)=(1)/(sqrt(g_(theta theta)))","quad(e_( hat(phi)))^(phi)=(1)/(sqrt(g_(phi phi))).]:}\begin{array}{lll} \left(\boldsymbol{e}_{t}\right)^{\hat{t}}=\sqrt{-g_{t t}}, & \left(\boldsymbol{e}_{r}\right)^{\hat{r}}=\sqrt{g_{r r}}, \quad\left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}}=\sqrt{g_{\theta \theta}}, \quad\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}}=\sqrt{g_{\phi \phi}}, \\ \left(\boldsymbol{e}_{\hat{t}}\right)^{t}=\frac{1}{\sqrt{-g_{t t}}}, \quad\left(\boldsymbol{e}_{\hat{r}}\right)^{r}=\frac{1}{\sqrt{g_{r r}}}, \quad\left(\boldsymbol{e}_{\hat{\theta}}\right)^{\theta}=\frac{1}{\sqrt{g_{\theta \theta}}}, \quad\left(\boldsymbol{e}_{\hat{\phi}}\right)^{\phi}=\frac{1}{\sqrt{g_{\phi \phi}}} . \end{array}(et)t^=gtt,(er)r^=grr,(eθ)θ^=gθθ,(eϕ)ϕ^=gϕϕ,(et^)t=1gtt,(er^)r=1grr,(eθ^)θ=1gθθ,(eϕ^)ϕ=1gϕϕ.
With this, we are finally ready to try some example computations.

36.4 Example computations

Here's the recipe for how to calculate the Riemann tensor components using these ideas (see also Fig. 36.1).
Step I: Identify the orthonormal basis from the metric, by comparing the given metric to
(36.42) d s 2 = g = ( ω t ^ ) 2 + ( ω χ ^ ) 2 + ( ω θ ^ ) 2 + ( ω ϕ ^ ) 2 . (36.42) d s 2 = g = ω t ^ 2 + ω χ ^ 2 + ω θ ^ 2 + ω ϕ ^ 2 . {:(36.42)ds^(2)=g=-(omega^( hat(t)))^(2)+(omega^( hat(chi)))^(2)+(omega^( hat(theta)))^(2)+(omega^( hat(phi)))^(2).:}\begin{equation*} \boldsymbol{d} \boldsymbol{s}^{2}=\boldsymbol{g}=-\left(\boldsymbol{\omega}^{\hat{t}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\chi}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\theta}}\right)^{2}+\left(\boldsymbol{\omega}^{\hat{\phi}}\right)^{2} . \tag{36.42} \end{equation*}(36.42)ds2=g=(ωt^)2+(ωχ^)2+(ωθ^)2+(ωϕ^)2.
In this basis, we have d g μ ν = d η μ ^ ν ^ = 0 d g μ ν = d η μ ^ ν ^ = 0 dg_(mu nu)=deta_( hat(mu) hat(nu))=0\boldsymbol{d} g_{\mu \nu}=\boldsymbol{d} \eta_{\hat{\mu} \hat{\nu}}=0dgμν=dημ^ν^=0, so, by the compatibility condition ω μ ^ ν ^ = ω ν ^ μ ^ ω μ ^ ν ^ = ω ν ^ μ ^ omega_( hat(mu) hat(nu))=-omega_( hat(nu) hat(mu))\boldsymbol{\omega}_{\hat{\mu} \hat{\nu}}=-\boldsymbol{\omega}_{\hat{\nu} \hat{\mu}}ωμ^ν^=ων^μ^.
Step II: Take the exterior derivatives of the basis 1 -forms. We have Idea 1 (the symmetry condition d ω μ ^ = ω μ ^ ν ^ ω ν ^ d ω μ ^ = ω μ ^ ν ^ ω ν ^ domega^( hat(mu))=-omega^( hat(mu))_( hat(nu))^^omega^( hat(nu))\boldsymbol{d} \boldsymbol{\omega}^{\hat{\mu}}=-\boldsymbol{\omega}^{\hat{\mu}}{ }_{\hat{\nu}} \wedge \boldsymbol{\omega}^{\hat{\nu}}dωμ^=ωμ^ν^ων^ ), and so it is an easy job to simply guess the values of ω μ ^ ν ^ ω μ ^ ν ^ omega^( hat(mu))_( hat(nu))\boldsymbol{\omega}^{\hat{\mu}}{ }_{\hat{\nu}}ωμ^ν^ from the results of the derivatives.
Step III: Take the exterior derivatives of the connection 1-forms ω μ ^ ν ^ ω μ ^ ν ^ omega^( hat(mu))_( hat(nu))\boldsymbol{\omega}^{\hat{\mu}}{ }_{\hat{\nu}}ωμ^ν^. Step IV: Assemble the curvature 2-form R μ ^ ν ^ R μ ^ ν ^ R^( hat(mu))_( hat(nu))\mathcal{R}^{\hat{\mu}}{ }_{\hat{\nu}}Rμ^ν^ using Idea 2.
Step V: Extract components R ν ^ α ^ β ^ μ ^ R ν ^ α ^ β ^ μ ^ R_( hat(nu) hat(alpha) hat(beta))^( hat(mu))R_{\hat{\nu} \hat{\alpha} \hat{\beta}}^{\hat{\mu}}Rν^α^β^μ^ as needed using Idea 3. Remember that you're currently working in the orthonormal frame, so it might be necessary to transform out of it using the vielbein components.
Let's discuss some examples. First we warm up with the (fairly trivial) case of flat two-dimensional space. Of course we expect the curvature
tensor to vanish if the space is truly flat. We won't include time in these first few computations, meaning that indices are raised and lowered in the orthonormal frame using the metric components η μ ν = diag ( 1 , 1 ) η μ ν = diag ( 1 , 1 ) eta^(mu nu)=diag(1,1)\eta^{\mu \nu}=\operatorname{diag}(1,1)ημν=diag(1,1), so that up and down become equivalent since this is Euclidean space.
Example 36.3
Flat space in polar coordinates has a metric line element d s 2 = d r 2 + r 2 d θ 2 d s 2 = d r 2 + r 2 d θ 2 ds^(2)=dr^(2)+r^(2)dtheta^(2)\boldsymbol{d} s^{2}=\boldsymbol{d} r^{2}+r^{2} \boldsymbol{d} \theta^{2}ds2=dr2+r2dθ2. We can identify an orthonormal basis with basis 1-forms
(36.45) ω r ^ = d r , ω θ ^ = r d θ (36.45) ω r ^ = d r , ω θ ^ = r d θ {:(36.45)omega^( hat(r))=dr","quadomega^( hat(theta))=rd theta:}\begin{equation*} \boldsymbol{\omega}^{\hat{r}}=\boldsymbol{d} r, \quad \boldsymbol{\omega}^{\hat{\theta}}=r \boldsymbol{d} \theta \tag{36.45} \end{equation*}(36.45)ωr^=dr,ωθ^=rdθ
The exterior derivatives of each of these are
(36.46) d ω r ^ = 0 , d ω θ ^ = d r d θ = ω θ ^ r ω r ^ (36.46) d ω r ^ = 0 , d ω θ ^ = d r d θ = ω θ ^ r ω r ^ {:(36.46)domega^( hat(r))=0","quad domega^( hat(theta))=dr^^d theta=-(omega^( hat(theta)))/(r)^^omega^( hat(r)):}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{r}}=0, \quad d \boldsymbol{\omega}^{\hat{\theta}}=\boldsymbol{d} r \wedge \boldsymbol{d} \theta=-\frac{\omega^{\hat{\theta}}}{r} \wedge \boldsymbol{\omega}^{\hat{r}} \tag{36.46} \end{equation*}(36.46)dωr^=0,dωθ^=drdθ=ωθ^rωr^
Using d ω θ ^ = ω θ ^ r ^ ω r ^ d ω θ ^ = ω θ ^ r ^ ω r ^ domega^( hat(theta))=-omega^( hat(theta))_( hat(r))^^omega^( hat(r))\boldsymbol{d} \boldsymbol{\omega}^{\hat{\theta}}=-\boldsymbol{\omega}^{\hat{\theta}}{ }_{\hat{r}} \wedge \boldsymbol{\omega}^{\hat{r}}dωθ^=ωθ^r^ωr^, we extract a non-zero connection 1-form
(36.47) ω r ^ θ ^ r ^ = ω θ ^ r = d θ (36.47) ω r ^ θ ^ r ^ = ω θ ^ r = d θ {:(36.47)omega_( hat(r))^( hat(theta)_( hat(r)))=(omega^( hat(theta)))/(r)=d theta:}\begin{equation*} \boldsymbol{\omega}_{\hat{r}}^{\hat{\theta}_{\hat{r}}}=\frac{\boldsymbol{\omega}^{\hat{\theta}}}{r}=\boldsymbol{d} \theta \tag{36.47} \end{equation*}(36.47)ωr^θ^r^=ωθ^r=dθ
Taking the exterior derivative of this connection 1-form, 11 11 ^(11){ }^{11}11 we find
(36.48) d ω r ^ θ ^ r ^ = d d θ = 0 (36.48) d ω r ^ θ ^ r ^ = d d θ = 0 {:(36.48)domega_( hat(r))^( hat(theta)_( hat(r)))=dd theta=0:}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}_{\hat{r}}^{\hat{\theta}_{\hat{r}}}=\boldsymbol{d} \boldsymbol{d} \theta=0 \tag{36.48} \end{equation*}(36.48)dωr^θ^r^=ddθ=0
As a result R μ ^ ν ^ = 0 R μ ^ ν ^ = 0 R^( hat(mu))_( hat(nu))=0\mathcal{R}^{\hat{\mu}}{ }_{\hat{\nu}}=0Rμ^ν^=0 for all components. This means that the Riemann tensor vanishes confirming (as we certainly knew) that this space is flat.
Next, we look at the space on the surface of the 2-sphere. Here we expect to see the effect of curvature.
Example 36.4
Spherical space has a metric line element d s 2 = a 2 d θ 2 + a 2 sin 2 θ d ϕ 2 d s 2 = a 2 d θ 2 + a 2 sin 2 θ d ϕ 2 ds^(2)=a^(2)dtheta^(2)+a^(2)sin^(2)theta dphi^(2)\boldsymbol{d} \boldsymbol{s}^{2}=a^{2} \boldsymbol{d} \theta^{2}+a^{2} \sin ^{2} \theta \boldsymbol{d} \phi^{2}ds2=a2dθ2+a2sin2θdϕ2, with a a aaa constant. We identify orthonormal basis 1 -forms
(36.51) ω θ ^ = a d θ , ω ϕ ^ = a sin θ d ϕ (36.51) ω θ ^ = a d θ , ω ϕ ^ = a sin θ d ϕ {:(36.51)omega^( hat(theta))=ad theta","quadomega^( hat(phi))=a sin theta d phi:}\begin{equation*} \boldsymbol{\omega}^{\hat{\theta}}=a \boldsymbol{d} \theta, \quad \boldsymbol{\omega}^{\hat{\phi}}=a \sin \theta \boldsymbol{d} \phi \tag{36.51} \end{equation*}(36.51)ωθ^=adθ,ωϕ^=asinθdϕ
We find exterior derivatives
(36.52) d ω θ ^ = 0 (36.52) d ω θ ^ = 0 {:(36.52)domega^( hat(theta))=0:}\begin{equation*} d \omega^{\hat{\theta}}=0 \tag{36.52} \end{equation*}(36.52)dωθ^=0
and
(36.53) d ω ϕ ^ = a cos d θ d ϕ = cos θ a sin θ ω ϕ ^ ω θ ^ (36.53) d ω ϕ ^ = a cos d θ d ϕ = cos θ a sin θ ω ϕ ^ ω θ ^ {:(36.53)domega^( hat(phi))=a cos d theta^^d phi=-(cos theta)/(a sin theta)omega^( hat(phi))^^omega^( hat(theta)):}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{\phi}}=a \cos \boldsymbol{d} \theta \wedge \boldsymbol{d} \phi=-\frac{\cos \theta}{a \sin \theta} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \tag{36.53} \end{equation*}(36.53)dωϕ^=acosdθdϕ=cosθasinθωϕ^ωθ^
We read off that a non-zero connection 1-form is given by
(36.54) ω θ ^ ϕ ^ = cos θ a sin θ ω ϕ ^ = cos θ d ϕ (36.54) ω θ ^ ϕ ^ = cos θ a sin θ ω ϕ ^ = cos θ d ϕ {:(36.54)omega_( hat(theta))^( hat(phi))=(cos theta)/(a sin theta)omega^( hat(phi))=cos theta d phi:}\begin{equation*} \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\phi}}=\frac{\cos \theta}{a \sin \theta} \boldsymbol{\omega}^{\hat{\phi}}=\cos \theta \boldsymbol{d} \phi \tag{36.54} \end{equation*}(36.54)ωθ^ϕ^=cosθasinθωϕ^=cosθdϕ
and, using ω i j ^ = ω j ^ i ^ ω i j ^ = ω j ^ i ^ omega_(i hat(j))=-omega_( hat(j) hat(i))\boldsymbol{\omega}_{i \hat{j}}=-\boldsymbol{\omega}_{\hat{j} \hat{i}}ωij^=ωj^i^, get for free that 12 12 ^(12){ }^{12}12
(36.56) ω ϕ ^ θ ^ = cos θ a sin θ ω ϕ ^ = cos θ d ϕ (36.56) ω ϕ ^ θ ^ = cos θ a sin θ ω ϕ ^ = cos θ d ϕ {:(36.56)omega_( hat(phi))^( hat(theta))=-(cos theta)/(a sin theta)*omega^( hat(phi))=-cos theta d phi:}\begin{equation*} \boldsymbol{\omega}_{\hat{\phi}}^{\hat{\theta}}=-\frac{\cos \theta}{a \sin \theta} \cdot \boldsymbol{\omega}^{\hat{\phi}}=-\cos \theta \boldsymbol{d} \phi \tag{36.56} \end{equation*}(36.56)ωϕ^θ^=cosθasinθωϕ^=cosθdϕ
Taking the exterior derivative of the non-zero connection 1-form yields
(36.57) d ω ϕ ^ θ ^ = sin θ d θ d ϕ = 1 a 2 ω θ ^ ω ϕ ^ (36.57) d ω ϕ ^ θ ^ = sin θ d θ d ϕ = 1 a 2 ω θ ^ ω ϕ ^ {:(36.57)domega_( hat(phi))^( hat(theta))=sin theta d theta^^d phi=(1)/(a^(2))omega^( hat(theta))^^omega^( hat(phi)):}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}_{\hat{\phi}}^{\hat{\theta}}=\sin \theta \boldsymbol{d} \theta \wedge \boldsymbol{d} \phi=\frac{1}{a^{2}} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \tag{36.57} \end{equation*}(36.57)dωϕ^θ^=sinθdθdϕ=1a2ωθ^ωϕ^
From this we can construct the non-zero components of the curvature 2-form. We find
(36.58) R ϕ ^ θ ^ = 1 a 2 ω θ ^ ω ϕ ^ (36.58) R ϕ ^ θ ^ = 1 a 2 ω θ ^ ω ϕ ^ {:(36.58)R_( hat(phi))^( hat(theta))=(1)/(a^(2))omega^( hat(theta))^^omega^( hat(phi)):}\begin{equation*} \mathcal{R}_{\hat{\phi}}^{\hat{\theta}}=\frac{1}{a^{2}} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \tag{36.58} \end{equation*}(36.58)Rϕ^θ^=1a2ωθ^ωϕ^
We have basis vectors
(36.43) e r ^ = r , e θ ^ = 1 r θ , (36.43) e r ^ = r , e θ ^ = 1 r θ , {:(36.43)e_( hat(r))=(del)/(del r)","quade_( hat(theta))=(1)/(r)(del)/(del theta)",":}\begin{equation*} e_{\hat{r}}=\frac{\partial}{\partial r}, \quad e_{\hat{\theta}}=\frac{1}{r} \frac{\partial}{\partial \theta}, \tag{36.43} \end{equation*}(36.43)er^=r,eθ^=1rθ,
and vielbein components
(36.44) ( e r ) r ^ = 1 , ( e θ ) θ ^ = r ( e r ^ ) r = 1 , ( e θ ^ ) θ = 1 / r (36.44) e r r ^ = 1 , e θ θ ^ = r e r ^ r = 1 , e θ ^ θ = 1 / r {:[(36.44)(e_(r))^( hat(r))=1","quad(e_(theta))^( hat(theta))=r],[(e_( hat(r)))^(r)=1","quad(e_( hat(theta)))^(theta)=1//r]:}\begin{align*} & \left(\boldsymbol{e}_{r}\right)^{\hat{r}}=1, \quad\left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}}=r \tag{36.44}\\ & \left(\boldsymbol{e}_{\hat{r}}\right)^{r}=1, \quad\left(\boldsymbol{e}_{\hat{\theta}}\right)^{\theta}=1 / r \end{align*}(36.44)(er)r^=1,(eθ)θ^=r(er^)r=1,(eθ^)θ=1/r
11 11 ^(11){ }^{11}11 Warning: To avoid making a mistake in taking this derivative, it's best to consider the version written in terms of the 1 -form d x μ d x μ dx^(mu)\boldsymbol{d} x^{\mu}dxμ rather than ω μ ω μ omega^(mu)\boldsymbol{\omega}^{\mu}ωμ. Note also that since ω i ^ j ^ = ω j ^ i ^ ω i ^ j ^ = ω j ^ i ^ omega_( hat(i) hat(j))=-omega_( hat(j) hat(i))\boldsymbol{\omega}_{\hat{i} \hat{j}}=-\boldsymbol{\omega}_{\hat{j} \hat{i}}ωi^j^=ωj^i^ (see Exercises) we have ω r ^ θ ^ = ω θ ^ θ ^ = ω θ / r ω r ^ θ ^ = ω θ ^ θ ^ = ω θ / r omega_( hat(r))^( hat(theta))=-omega_( hat(theta))^( hat(theta))=omega^(theta)//r\boldsymbol{\omega}_{\hat{r}}^{\hat{\theta}}=-\boldsymbol{\omega}_{\hat{\theta}}^{\hat{\theta}}=\boldsymbol{\omega}^{\theta} / rωr^θ^=ωθ^θ^=ωθ/r, so that ω r ^ θ ^ ω θ ^ θ ^ = 0 ω r ^ θ ^ ω θ ^ θ ^ = 0 omega_( hat(r))^( hat(theta))^^omega_( hat(theta))^( hat(theta))=0\boldsymbol{\omega}_{\hat{r}}^{\hat{\theta}} \wedge \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\theta}}=0ωr^θ^ωθ^θ^=0.
Basis vectors:
(36.49) e θ ^ = 1 a θ , e ϕ ^ = 1 a sin θ ϕ , (36.49) e θ ^ = 1 a θ , e ϕ ^ = 1 a sin θ ϕ , {:(36.49)e_( hat(theta))=(1)/(a)(del)/(del theta)","quade_( hat(phi))=(1)/(a sin theta)(del)/(del phi)",":}\begin{equation*} \boldsymbol{e}_{\hat{\theta}}=\frac{1}{a} \frac{\partial}{\partial \theta}, \quad \boldsymbol{e}_{\hat{\phi}}=\frac{1}{a \sin \theta} \frac{\partial}{\partial \phi}, \tag{36.49} \end{equation*}(36.49)eθ^=1aθ,eϕ^=1asinθϕ,
and vielbein components:
( e θ ) θ ^ = a , ( e ϕ ) ϕ ^ = a sin θ ( e θ ^ ) θ = 1 a , ( e ϕ ^ ) ϕ = 1 a sin θ . ( 36.50 ) e θ θ ^ = a , e ϕ ϕ ^ = a sin θ e θ ^ θ = 1 a , e ϕ ^ ϕ = 1 a sin θ . ( 36.50 ) {:[(e_(theta))^( hat(theta))=a",",(e_(phi))^( hat(phi))=a sin theta],[(e_( hat(theta)))^(theta)=(1)/(a)",",(e_( hat(phi)))^(phi)=(1)/(a sin theta).],[(36.50)]:}\begin{array}{ll} \left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}}=a, & \left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}}=a \sin \theta \\ \left(\boldsymbol{e}_{\hat{\theta}}\right)^{\theta}=\frac{1}{a}, & \left(\boldsymbol{e}_{\hat{\phi}}\right)^{\phi}=\frac{1}{a \sin \theta} . \\ (36.50) \end{array}(eθ)θ^=a,(eϕ)ϕ^=asinθ(eθ^)θ=1a,(eϕ^)ϕ=1asinθ.(36.50)
12 12 ^(12){ }^{12}12 We could at this point extract connection coefficients
Γ ϕ ˙ ϕ ^ θ ^ = 1 a cot θ , (36.55) Γ ϕ ˙ ϕ θ ^ = 1 a cot θ . Γ ϕ ˙ ϕ ^ θ ^ = 1 a cot θ , (36.55) Γ ϕ ˙ ϕ θ ^ = 1 a cot θ . {:[Gamma_(phi^(˙) hat(phi))^( hat(theta))=-(1)/(a)cot theta","],[(36.55)Gamma^(phi^(˙))_(phi hat(theta))=(1)/(a)cot theta.]:}\begin{align*} \Gamma_{\dot{\phi} \hat{\phi}}^{\hat{\theta}} & =-\frac{1}{a} \cot \theta, \\ \Gamma^{\dot{\phi}}{ }_{\phi \hat{\theta}} & =\frac{1}{a} \cot \theta . \tag{36.55} \end{align*}Γϕ˙ϕ^θ^=1acotθ,(36.55)Γϕ˙ϕθ^=1acotθ.
3 3 ^(3){ }^{3}3 Here the symbol | α β | | α β | |alpha beta||\alpha \beta||αβ| orders components in the order they appear in the wedge product. That is, | α ^ β ^ | = θ ^ ϕ ^ | α ^ β ^ | = θ ^ ϕ ^ | hat(alpha) hat(beta)|= hat(theta) hat(phi)|\hat{\alpha} \hat{\beta}|=\hat{\theta} \hat{\phi}|α^β^|=θ^ϕ^.
Using 13 13 ^(13){ }^{13}13
we read off
(36.59) R ϕ ^ θ ^ = R ϕ ^ | θ ^ ϕ ^ | θ ^ ω θ ^ ω ϕ ^ , (36.59) R ϕ ^ θ ^ = R ϕ ^ | θ ^ ϕ ^ | θ ^ ω θ ^ ω ϕ ^ , {:(36.59)R_( hat(phi))^( hat(theta))=R_( hat(phi)| hat(theta) hat(phi)|)^( hat(theta))omega^( hat(theta))^^omega^( hat(phi))",":}\begin{equation*} \mathcal{R}_{\hat{\phi}}^{\hat{\theta}}=R_{\hat{\phi}|\hat{\theta} \hat{\phi}|}^{\hat{\theta}} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\phi}}, \tag{36.59} \end{equation*}(36.59)Rϕ^θ^=Rϕ^|θ^ϕ^|θ^ωθ^ωϕ^,
Note that in the coordinate frame
R ϕ ^ θ ^ ϕ ^ θ ^ = 1 a 2 . R ϕ ^ θ ^ ϕ ^ θ ^ = 1 a 2 . R_( hat(phi) hat(theta) hat(phi))^( hat(theta))=(1)/(a^(2)).R_{\hat{\phi} \hat{\theta} \hat{\phi}}^{\hat{\theta}}=\frac{1}{a^{2}} .Rϕ^θ^ϕ^θ^=1a2.
R ϕ θ ϕ θ = R ϕ ^ θ ^ ϕ ^ θ ^ ( e θ ^ ) θ ( e ϕ ) ϕ ^ ( e θ ) θ ^ ( e ϕ ) ϕ ^ (36.61) = 1 a 2 a 2 sin 2 θ = sin 2 θ . R ϕ θ ϕ θ = R ϕ ^ θ ^ ϕ ^ θ ^ e θ ^ θ e ϕ ϕ ^ e θ θ ^ e ϕ ϕ ^ (36.61) = 1 a 2 a 2 sin 2 θ = sin 2 θ . {:[R_(phi theta phi)^(theta)=R_( hat(phi) hat(theta) hat(phi))^( hat(theta))(e_( hat(theta)))^(theta)(e_(phi))^( hat(phi))(e_(theta))^( hat(theta))(e_(phi))^( hat(phi))],[(36.61)=(1)/(a^(2))a^(2)sin^(2)theta=sin^(2)theta.]:}\begin{align*} R_{\phi \theta \phi}^{\theta} & =R_{\hat{\phi} \hat{\theta} \hat{\phi}}^{\hat{\theta}}\left(\boldsymbol{e}_{\hat{\theta}}\right)^{\theta}\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}}\left(\boldsymbol{e}_{\theta}\right)^{\hat{\theta}}\left(\boldsymbol{e}_{\phi}\right)^{\hat{\phi}} \\ & =\frac{1}{a^{2}} a^{2} \sin ^{2} \theta=\sin ^{2} \theta . \tag{36.61} \end{align*}Rϕθϕθ=Rϕ^θ^ϕ^θ^(eθ^)θ(eϕ)ϕ^(eθ)θ^(eϕ)ϕ^(36.61)=1a2a2sin2θ=sin2θ.
14 14 ^(14){ }^{14}14 The easiest way to proceed here is We also find 14 14 ^(14){ }^{14}14 to manipulate indices using the symmetries of R R R\boldsymbol{R}R. In this case,
(36.62) 1 a 2 = R ϕ ^ θ ^ ϕ ^ θ ^ = R θ ^ ϕ ^ θ ^ ϕ ^ . (36.62) 1 a 2 = R ϕ ^ θ ^ ϕ ^ θ ^ = R θ ^ ϕ ^ θ ^ ϕ ^ . {:(36.62)(1)/(a^(2))=R_( hat(phi) hat(theta) hat(phi))^( hat(theta))=R_( hat(theta) hat(phi) hat(theta) hat(phi)).:}\begin{equation*} \frac{1}{a^{2}}=R_{\hat{\phi} \hat{\theta} \hat{\phi}}^{\hat{\theta}}=R_{\hat{\theta} \hat{\phi} \hat{\theta} \hat{\phi}} . \tag{36.62} \end{equation*}(36.62)1a2=Rϕ^θ^ϕ^θ^=Rθ^ϕ^θ^ϕ^.
Recall that a swap within the two pairs of indices pick up a minus sign, so
(36.63) R θ ^ ϕ ^ θ ^ ϕ ^ = R ϕ ^ θ ^ ϕ ^ θ ^ , (36.63) R θ ^ ϕ ^ θ ^ ϕ ^ = R ϕ ^ θ ^ ϕ ^ θ ^ , {:(36.63)R_( hat(theta) hat(phi) hat(theta) hat(phi))=R_( hat(phi) hat(theta) hat(phi) hat(theta))",":}\begin{equation*} R_{\hat{\theta} \hat{\phi} \hat{\theta} \hat{\phi}}=R_{\hat{\phi} \hat{\theta} \hat{\phi} \hat{\theta}}, \tag{36.63} \end{equation*}(36.63)Rθ^ϕ^θ^ϕ^=Rϕ^θ^ϕ^θ^,
and finally, since up and down are equivalent for spatial indices
(36.64) R θ ^ ϕ ^ θ ^ ϕ ^ = 1 a 2 . (36.64) R θ ^ ϕ ^ θ ^ ϕ ^ = 1 a 2 . {:(36.64)R_( hat(theta) hat(phi) hat(theta))^( hat(phi))=(1)/(a^(2)).:}\begin{equation*} R_{\hat{\theta} \hat{\phi} \hat{\theta}}^{\hat{\phi}}=\frac{1}{a^{2}} . \tag{36.64} \end{equation*}(36.64)Rθ^ϕ^θ^ϕ^=1a2.
15 15 ^(15){ }^{15}15 We called this Universe 1 in Chapter 15 .
Basis vectors
(36.70) e t ^ = t , (36.71) e k ^ = 1 a ( t ) x k . (36.70) e t ^ = t , (36.71) e k ^ = 1 a ( t ) x k . {:[(36.70)e_( hat(t))=(del)/(del t)","],[(36.71)e_( hat(k))=(1)/(a(t))*(del)/(delx^(k)).]:}\begin{align*} & \boldsymbol{e}_{\hat{t}}=\frac{\partial}{\partial t}, \tag{36.70}\\ & \boldsymbol{e}_{\hat{k}}=\frac{1}{a(t)} \cdot \frac{\partial}{\partial x^{k}} . \tag{36.71} \end{align*}(36.70)et^=t,(36.71)ek^=1a(t)xk.
Vielbein components
(36.72) ( e t ) t ^ = 1 , ( e k ) k ^ = a ( t ) ( e t ^ ) t = 1 , ( e k ^ ) k = 1 a ( t ) (36.72) e t t ^ = 1 , e k k ^ = a ( t ) e t ^ t = 1 , e k ^ k = 1 a ( t ) {:(36.72){:[(e_(t))^( hat(t))=1",",(e_(k))^( hat(k))=a(t)],[(e_( hat(t)))^(t)=1",",(e_( hat(k)))^(k)=(1)/(a(t))]:}:}\begin{array}{ll} \left(\boldsymbol{e}_{t}\right)^{\hat{t}}=1, & \left(\boldsymbol{e}_{k}\right)^{\hat{k}}=a(t) \tag{36.72}\\ \left(\boldsymbol{e}_{\hat{t}}\right)^{t}=1, & \left(\boldsymbol{e}_{\hat{k}}\right)^{k}=\frac{1}{a(t)} \end{array}(36.72)(et)t^=1,(ek)k^=a(t)(et^)t=1,(ek^)k=1a(t)
16 16 ^(16){ }^{16}16 Useful here is to recall that raising a lowering time indices in the orthonormal frame picks up a minus sign, so we have
(36.75) ω t ^ k ^ = ω k ^ t ^ = ω t ^ k ^ = ω k ^ t ^ . (36.75) ω t ^ k ^ = ω k ^ t ^ = ω t ^ k ^ = ω k ^ t ^ . {:(36.75)omega_( hat(t))^( hat(k))=omega_( hat(k) hat(t))=-omega_( hat(t) hat(k))=omega_( hat(k))^( hat(t)).:}\begin{equation*} \boldsymbol{\omega}_{\hat{t}}^{\hat{k}}=\boldsymbol{\omega}_{\hat{k} \hat{t}}=-\boldsymbol{\omega}_{\hat{t} \hat{k}}=\boldsymbol{\omega}_{\hat{k}}^{\hat{t}} . \tag{36.75} \end{equation*}(36.75)ωt^k^=ωk^t^=ωt^k^=ωk^t^.
which, in the cordinate frame, R θ ^ ϕ ^ θ ^ θ ˙ = 1 a 2 R θ ^ ϕ ^ θ ^ θ ˙ = 1 a 2 R_( hat(theta) hat(phi) hat(theta))^(theta^(˙))=(1)/(a^(2))R_{\hat{\theta} \hat{\phi} \hat{\theta}}^{\dot{\theta}}=\frac{1}{a^{2}}Rθ^ϕ^θ^θ˙=1a2,
which, in the coordinate frame, becomes
(36.66) R θ ϕ θ ϕ = 1 a 2 ( e θ ) θ ^ ( e θ ) θ ^ = 1 (36.66) R θ ϕ θ ϕ = 1 a 2 e θ θ ^ e θ θ ^ = 1 {:(36.66)R_(theta phi theta)^(phi)=(1)/(a^(2))(e_(theta))^( hat(theta))(e_(theta))^( hat(theta))=1:}\begin{equation*} R_{\theta \phi \theta}^{\phi}=\frac{1}{a^{2}}\left(e_{\theta}\right)^{\hat{\theta}}\left(e_{\theta}\right)^{\hat{\theta}}=1 \tag{36.66} \end{equation*}(36.66)Rθϕθϕ=1a2(eθ)θ^(eθ)θ^=1
The components of the Ricci tensor calculated in the orthonormal frame, are
(36.67) R ϕ ^ ϕ ^ = R θ ^ θ ^ = 1 a 2 (36.67) R ϕ ^ ϕ ^ = R θ ^ θ ^ = 1 a 2 {:(36.67)R_( hat(phi) hat(phi))=R_( hat(theta) hat(theta))=(1)/(a^(2)):}\begin{equation*} R_{\hat{\phi} \hat{\phi}}=R_{\hat{\theta} \hat{\theta}}=\frac{1}{a^{2}} \tag{36.67} \end{equation*}(36.67)Rϕ^ϕ^=Rθ^θ^=1a2
and so, finally, the Ricci scalar is
(36.68) R = 2 a 2 (36.68) R = 2 a 2 {:(36.68)R=(2)/(a^(2)):}\begin{equation*} R=\frac{2}{a^{2}} \tag{36.68} \end{equation*}(36.68)R=2a2
Note that, being a scalar, R R RRR is the same if we choose to evaluate it in the coordinate frame using the Ricci tensor components R μ ν R μ ν R_(mu nu)R_{\mu \nu}Rμν.
What can be done on the surface of a sphere can be done on other surfaces and there are several examples in the exercises at the end of the chapter. Let's now include time and find the curvature of some nontrivial spacetimes. First is an expanding Universe model from Chapter 18 where space is flat, but spacetime is not.

Example 36.5

The de Sitter model of the Universe 15 15 ^(15){ }^{15}15 has a metric with line element
(36.69) d s 2 = d t 2 + a ( t ) 2 ( d x 2 + d y 2 + d z 2 ) (36.69) d s 2 = d t 2 + a ( t ) 2 d x 2 + d y 2 + d z 2 {:(36.69)ds^(2)=-dt^(2)+a(t)^(2)(dx^(2)+dy^(2)+dz^(2)):}\begin{equation*} \boldsymbol{d} s^{2}=-\boldsymbol{d} t^{2}+a(t)^{2}\left(\boldsymbol{d} x^{2}+\boldsymbol{d} y^{2}+\boldsymbol{d} z^{2}\right) \tag{36.69} \end{equation*}(36.69)ds2=dt2+a(t)2(dx2+dy2+dz2)
We have orthonormal basis 1-forms
ω t ^ = d t , ω k ^ = a ( t ) d x k d ω t ^ = 0 ω t ^ = d t , ω k ^ = a ( t ) d x k d ω t ^ = 0 {:[omega^( hat(t))=dt","quadomega^( hat(k))=a(t)dx^(k)],[domega^( hat(t))=0]:}\begin{gathered} \boldsymbol{\omega}^{\hat{t}}=\boldsymbol{d} t, \quad \boldsymbol{\omega}^{\hat{k}}=a(t) \boldsymbol{d} x^{k} \\ \boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}=0 \end{gathered}ωt^=dt,ωk^=a(t)dxkdωt^=0
First we note
and we use d ω t ^ = ω t ^ k ^ ω k ^ d ω t ^ = ω t ^ k ^ ω k ^ domega^( hat(t))=-omega^( hat(t))_( hat(k))^^omega^( hat(k))\boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}=-\boldsymbol{\omega}^{\hat{t}}{ }_{\hat{k}} \wedge \boldsymbol{\omega}^{\hat{k}}dωt^=ωt^k^ωk^, from which we guess that ω k ^ t ^ ω k ^ ω k ^ t ^ ω k ^ omega_( hat(k))^( hat(t))propomega^( hat(k))\boldsymbol{\omega}_{\hat{k}}^{\hat{t}} \propto \boldsymbol{\omega}^{\hat{k}}ωk^t^ωk^. Then
d ω k ^ = a ˙ a ω t ^ ω k ^ d ω k ^ = a ˙ a ω t ^ ω k ^ domega^( hat(k))=((a^(˙)))/(a)*omega^( hat(t))^^omega^( hat(k))\boldsymbol{d} \boldsymbol{\omega}^{\hat{k}}=\frac{\dot{a}}{a} \cdot \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{k}}dωk^=a˙aωt^ωk^
Writing
(36.73) d ω k ^ = ω t ^ k ^ ω t ^ ω i k ^ ω i ^ , (36.73) d ω k ^ = ω t ^ k ^ ω t ^ ω i k ^ ω i ^ , {:(36.73)domega^( hat(k))=-omega_( hat(t))^( hat(k))^^omega^( hat(t))-omega_(i)^( hat(k))^^omega^( hat(i))",":}\begin{equation*} \boldsymbol{d} \boldsymbol{\omega}^{\hat{k}}=-\boldsymbol{\omega}_{\hat{t}}^{\hat{k}} \wedge \boldsymbol{\omega}^{\hat{t}}-\boldsymbol{\omega}_{i}^{\hat{k}} \wedge \boldsymbol{\omega}^{\hat{i}}, \tag{36.73} \end{equation*}(36.73)dωk^=ωt^k^ωt^ωik^ωi^,
we guess ω i ^ k ^ = a ˙ a ω k ^ ω i ^ k ^ = a ˙ a ω k ^ omega_( hat(i))^( hat(k))=((a^(˙)))/(a)omega^( hat(k))\boldsymbol{\omega}_{\hat{i}}^{\hat{k}}=\frac{\dot{a}}{a} \omega^{\hat{k}}ωi^k^=a˙aωk^ and ω l ^ k ^ = 0 ω l ^ k ^ = 0 omega_( hat(l))^( hat(k))=0\boldsymbol{\omega}_{\hat{l}}^{\hat{k}}=0ωl^k^=0. Now to compute the curvature 2-form
(36.74) R k ^ k ^ = d ω t ^ k ^ = d ( a ˙ d x k ^ ) = a ¨ a ω t ^ ω k ^ (36.74) R k ^ k ^ = d ω t ^ k ^ = d a ˙ d x k ^ = a ¨ a ω t ^ ω k ^ {:(36.74)R^( hat(k))_( hat(k))=domega^( hat(t))_( hat(k))=d((a^(˙))dx^( hat(k)))=((a^(¨)))/(a)omega^( hat(t))^^omega^( hat(k)):}\begin{equation*} \mathcal{R}^{\hat{k}}{ }_{\hat{k}}=\boldsymbol{d} \boldsymbol{\omega}^{\hat{t}}{ }_{\hat{k}}=\boldsymbol{d}\left(\dot{a} \boldsymbol{d} x^{\hat{k}}\right)=\frac{\ddot{a}}{a} \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{k}} \tag{36.74} \end{equation*}(36.74)Rk^k^=dωt^k^=d(a˙dxk^)=a¨aωt^ωk^
We must also not forget that there are other components in this case. We have 16 16 ^(16){ }^{16}16
R i k ^ = d ω i k ^ + ω k ^ i ω i t ^ (36.76) = 0 + ( a ˙ a ) 2 ω k ^ ω l ^ . R i k ^ = d ω i k ^ + ω k ^ i ω i t ^ (36.76) = 0 + a ˙ a 2 ω k ^ ω l ^ . {:[R_(i)^( hat(k))=domega_(i)^( hat(k))+omega^( hat(k))_(i)^^omega_(i)^( hat(t))],[(36.76)=0+(((a^(˙)))/(a))^(2)omega^( hat(k))^^omega^( hat(l)).]:}\begin{align*} \mathcal{R}_{i}^{\hat{k}} & =\boldsymbol{d} \boldsymbol{\omega}_{i}^{\hat{k}}+\boldsymbol{\omega}^{\hat{k}}{ }_{i} \wedge \boldsymbol{\omega}_{i}^{\hat{t}} \\ & =0+\left(\frac{\dot{a}}{a}\right)^{2} \boldsymbol{\omega}^{\hat{k}} \wedge \boldsymbol{\omega}^{\hat{l}} . \tag{36.76} \end{align*}Rik^=dωik^+ωk^iωit^(36.76)=0+(a˙a)2ωk^ωl^.
We obtain the resulting Riemann tensor components 17 17 ^(17){ }^{17}17
R k ^ t k ^ t ^ = a ¨ a , R i k ^ l ^ k ^ = ( a ˙ a ˙ ) 2 R k ^ t k ^ t ^ = a ¨ a , R i k ^ l ^ k ^ = a ˙ a ˙ 2 R_( hat(k)t hat(k))^( hat(t))=((a^(¨)))/(a),quadR_(i hat(k) hat(l))^( hat(k))=(((a^(˙)))/((a^(˙))))^(2)R_{\hat{k} t \hat{k}}^{\hat{t}}=\frac{\ddot{a}}{a}, \quad R_{i \hat{k} \hat{l}}^{\hat{k}}=\left(\frac{\dot{a}}{\dot{a}}\right)^{2}Rk^tk^t^=a¨a,Rik^l^k^=(a˙a˙)2
(36.77)
where we have temporarily suspended the Einstein summation convention. The Ricci tensor has 0̂ôth component
(36.78) R i t = k = 1 3 R k t k ¯ t = 3 a ¨ a (36.78) R i t = k = 1 3 R k t k ¯ t = 3 a ¨ a {:(36.78)R_(it)=sum_(k=1)^(3)R^(k)_(t bar(k)t)=-3((a^(¨)))/(a):}\begin{equation*} R_{i t}=\sum_{k=1}^{3} R^{k}{ }_{t \bar{k} t}=-3 \frac{\ddot{a}}{a} \tag{36.78} \end{equation*}(36.78)Rit=k=13Rktk¯t=3a¨a
To find the i ^ i ^ i ^ i ^ hat(i) hat(i)\hat{i} \hat{i}i^i^ th components we have R i ^ i ^ = R t ^ i i ^ i ^ ^ + j = 1 3 R i ^ j ^ i ^ i ^ j ^ R i ^ i ^ = R t ^ i i ^ i ^ ^ + j = 1 3 R i ^ j ^ i ^ i ^ j ^ R_( hat(i) hat(i))=R^( hat(t)) hat(i( hat(i))( hat(i)))+sum_(j=1)^(3)R^( hat(i) hat(j) hat(i)) hat(i)^( hat(j))R_{\hat{i} \hat{i}}=R^{\hat{t}} \hat{i \hat{i} \hat{i}}+\sum_{j=1}^{3} R^{\hat{i} \hat{j} \hat{i}} \hat{i}^{\hat{j}}Ri^i^=Rt^ii^i^^+j=13Ri^j^i^i^j^. However, the component with i ^ = j ^ i ^ = j ^ hat(i)= hat(j)\hat{i}=\hat{j}i^=j^ is R i ^ i ^ i ^ = 0 R i ^ i ^ i ^ = 0 R_( hat(i) hat(i) hat(i))=0R_{\hat{i} \hat{i} \hat{i}}=0Ri^i^i^=0 because of the symmetries of the Riemann tensor. As a result we have
(36.79) R i ¯ i ¯ = a ¨ a + 2 ( a ˙ a ) 2 (36.79) R i ¯ i ¯ = a ¨ a + 2 a ˙ a 2 {:(36.79)R_( bar(i) bar(i))=((a^(¨)))/(a)+2(((a^(˙)))/(a))^(2):}\begin{equation*} R_{\bar{i} \bar{i}}=\frac{\ddot{a}}{a}+2\left(\frac{\dot{a}}{a}\right)^{2} \tag{36.79} \end{equation*}(36.79)Ri¯i¯=a¨a+2(a˙a)2
The Ricci scalar is therefore
(36.80) R = 3 a ¨ a + 3 [ a ¨ a + 2 ( a ˙ a ) 2 ] = 6 ( a ¨ a + a ˙ 2 a 2 ) (36.80) R = 3 a ¨ a + 3 a ¨ a + 2 a ˙ a 2 = 6 a ¨ a + a ˙ 2 a 2 {:(36.80)R=3((a^(¨)))/(a)+3[((a^(¨)))/(a)+2(((a^(˙)))/(a))^(2)]=6(((a^(¨)))/(a)+(a^(˙)^(2))/(a^(2))):}\begin{equation*} R=3 \frac{\ddot{a}}{a}+3\left[\frac{\ddot{a}}{a}+2\left(\frac{\dot{a}}{a}\right)^{2}\right]=6\left(\frac{\ddot{a}}{a}+\frac{\dot{a}^{2}}{a^{2}}\right) \tag{36.80} \end{equation*}(36.80)R=3a¨a+3[a¨a+2(a˙a)2]=6(a¨a+a˙2a2)
In this case, it's meaningful to extract the components of the Einstein tensor G μ ν = G μ ν = G_(mu nu)=G_{\mu \nu}=Gμν= R μ ν 1 2 g μ ν R R μ ν 1 2 g μ ν R R_(mu nu)-(1)/(2)g_(mu nu)RR_{\mu \nu}-\frac{1}{2} g_{\mu \nu} RRμν12gμνR. These are
(36.81) G i t = 3 ( a ¨ a ) 2 , G i i ^ = 2 a ¨ a a ˙ a (36.81) G i t = 3 a ¨ a 2 , G i i ^ = 2 a ¨ a a ˙ a {:(36.81)G_(it)=3(((a^(¨)))/(a))^(2)","quadG_(i hat(i))=-2((a^(¨)))/(a)-((a^(˙)))/(a):}\begin{equation*} G_{i t}=3\left(\frac{\ddot{a}}{a}\right)^{2}, \quad G_{i \hat{i}}=-2 \frac{\ddot{a}}{a}-\frac{\dot{a}}{a} \tag{36.81} \end{equation*}(36.81)Git=3(a¨a)2,Gii^=2a¨aa˙a
Finally, we can compute the curvature properties of the Schwarzschild metric from Chapter 21.

Example 36.6

Written in its general form, the spherically symmetric metric line element looks like
\begin{align*} & \qquad \boldsymbol{d} \boldsymbol{s}^{2}=-\mathrm{e}^{2 \Phi} \boldsymbol{d} t^{2}+\mathrm{e}^{2 \Lambda} \boldsymbol{d}^{2}+r^{2}\left(\boldsymbol{d} \theta^{2}+\sin ^{2} \theta \boldsymbol{d} \phi^{2}\right), \tag{36.82}\\ & \text { where } \Phi \text { and } \Lambda \text { are functions of } r \text { only. We identify orthonormal basis 1-forms } \\ & \qquad \boldsymbol{\omega}^{\hat{t}}=\mathrm{e}^{\Phi} \boldsymbol{d} t, \quad \boldsymbol{\omega}^{\hat{r}}=\mathrm{e}^{\Lambda} \boldsymbol{d} r, \quad \boldsymbol{\omega}^{\hat{\theta}}=r \boldsymbol{d} \boldsymbol{\theta} \theta, \quad \boldsymbol{\omega}^{\hat{\phi}}=r \sin \theta \boldsymbol{d} \phi \end{aligned} \begin{aligned} & \text { The connections are } \tag{36.83}\\ & \qquad \begin{array}{l} \boldsymbol{\omega}_{\hat{t}}^{\hat{t}}=\Phi^{\prime} \mathrm{e}^{\Phi} \mathrm{e}^{-\Lambda} \boldsymbol{d} t, \quad \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\theta}}=\mathrm{e}^{-\Lambda} \boldsymbol{d} \theta, \\ \boldsymbol{\omega}_{\hat{r}}^{\phi}=\sin \theta \mathrm{e}^{-\Lambda} \boldsymbol{d} \phi, \quad \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\phi}}=\cos \theta \boldsymbol{d} \phi, \end{array} \\ & \text { where dashes denote derivatives with respect to } r \text {. The exterior derivatives of the } \tag{36.84}\\ & \text { connections can be computed and pive } \end{align*}\begin{align*} ended with \end{aligned} connections can be computed and give
d ω r ^ t ^ = [ Φ + ( Φ ) 2 Φ Λ ] e 2 Λ ω t ^ ω θ ^ d ω r ^ θ ^ = Λ e 2 Λ r ω θ ^ ω r ^ d ω r ^ ϕ ^ = ( e Λ cos θ r 2 sin θ + Λ e 2 Λ r ) ω ϕ ^ ω θ ^ (36.85) d ω θ ^ ϕ ^ = 1 r 2 ω ϕ ^ ω θ ^ d ω r ^ t ^ = Φ + Φ 2 Φ Λ e 2 Λ ω t ^ ω θ ^ d ω r ^ θ ^ = Λ e 2 Λ r ω θ ^ ω r ^ d ω r ^ ϕ ^ = e Λ cos θ r 2 sin θ + Λ e 2 Λ r ω ϕ ^ ω θ ^ (36.85) d ω θ ^ ϕ ^ = 1 r 2 ω ϕ ^ ω θ ^ {:[domega_( hat(r))^( hat(t))=-[Phi^('')+(Phi^('))^(2)-Phi^(')Lambda^(')]e^(-2Lambda)omega^( hat(t))^^omega^( hat(theta))],[domega_( hat(r))^( hat(theta))=(Lambda^(')e^(-2Lambda))/(r)omega^( hat(theta))^^omega^( hat(r))],[domega_( hat(r))^( hat(phi))=(-(e^(-Lambda)cos theta)/(r^(2)sin theta)+(Lambdae^(-2Lambda))/(r))omega^( hat(phi))^^omega^( hat(theta))],[(36.85)domega_( hat(theta))^( hat(phi))=(1)/(r^(2))omega^( hat(phi))^^omega^( hat(theta))]:}\begin{align*} & \boldsymbol{d} \boldsymbol{\omega}_{\hat{r}}^{\hat{t}}=-\left[\Phi^{\prime \prime}+\left(\Phi^{\prime}\right)^{2}-\Phi^{\prime} \Lambda^{\prime}\right] \mathrm{e}^{-2 \Lambda} \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \\ & \boldsymbol{d} \boldsymbol{\omega}_{\hat{r}}^{\hat{\theta}}=\frac{\Lambda^{\prime} \mathrm{e}^{-2 \Lambda}}{r} \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{r}} \\ & \boldsymbol{d} \boldsymbol{\omega}_{\hat{r}}^{\hat{\phi}}=\left(-\frac{\mathrm{e}^{-\Lambda} \cos \theta}{r^{2} \sin \theta}+\frac{\Lambda \mathrm{e}^{-2 \Lambda}}{r}\right) \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \\ & \boldsymbol{d} \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\phi}}=\frac{1}{r^{2}} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \tag{36.85} \end{align*}dωr^t^=[Φ+(Φ)2ΦΛ]e2Λωt^ωθ^dωr^θ^=Λe2Λrωθ^ωr^dωr^ϕ^=(eΛcosθr2sinθ+Λe2Λr)ωϕ^ωθ^(36.85)dωθ^ϕ^=1r2ωϕ^ωθ^
Next, we extract the non-zero components of the curvature 2-form
R r ^ r ^ = d ω r ^ , R θ ^ i ^ = ω r ^ t ^ ω θ ^ ^ ^ ^ , R t ϕ = ω t ^ r ^ ω r ^ ϕ ^ , R ϕ ^ θ ^ = d ω ϕ ^ θ ^ + ω θ ^ θ ^ ω ϕ ^ ϕ ^ , R r ^ ϕ ^ = d ω r ^ ϕ ^ + ω θ ^ ϕ ^ ω r ^ θ ^ , R θ ^ r ^ = d ω θ ^ r ^ + ω ϕ ^ r ^ ω θ ^ ϕ ^ . R r ^ r ^ = d ω r ^ , R θ ^ i ^ = ω r ^ t ^ ω θ ^ ^ ^ ^ R t ϕ = ω t ^ r ^ ω r ^ ϕ ^ , R ϕ ^ θ ^ = d ω ϕ ^ θ ^ + ω θ ^ θ ^ ω ϕ ^ ϕ ^ , R r ^ ϕ ^ = d ω r ^ ϕ ^ + ω θ ^ ϕ ^ ω r ^ θ ^ , R θ ^ r ^ = d ω θ ^ r ^ + ω ϕ ^ r ^ ω θ ^ ϕ ^ . {:[R_( hat(r))_( hat(r))=domega_( hat(r))","],[R_( hat(theta))^( hat(i))=omega_( hat(r))^( hat(t))^^omega_( hat(hat(theta)))^( hat(hat()))^(", ")],[R^(t)_(phi)=omega^( hat(t))_( hat(r))^^omega^( hat(r))_( hat(phi))","],[R_( hat(phi))^( hat(theta))=domega_( hat(phi))^( hat(theta))+omega_( hat(theta))^( hat(theta))^^omega_( hat(phi))^( hat(phi))","],[R_( hat(r))^( hat(phi))=domega_( hat(r))^( hat(phi))+omega_( hat(theta))^( hat(phi))^^omega_( hat(r))^( hat(theta))","],[R_( hat(theta))^( hat(r))=domega_( hat(theta))^( hat(r))+omega_( hat(phi))^( hat(r))^^omega_( hat(theta))^( hat(phi)).]:}\begin{aligned} & \mathcal{R}_{\hat{r}}{ }_{\hat{r}}=\boldsymbol{d} \boldsymbol{\omega}_{\hat{r}}, \\ & \mathcal{R}_{\hat{\theta}}^{\hat{i}}=\boldsymbol{\omega}_{\hat{r}}^{\hat{t}} \wedge \boldsymbol{\omega}_{\hat{\hat{\theta}}}^{\hat{\hat{}}}{ }^{\text {, }} \\ & \mathcal{R}^{t}{ }_{\phi}=\boldsymbol{\omega}^{\hat{t}}{ }_{\hat{r}} \wedge \boldsymbol{\omega}^{\hat{r}}{ }_{\hat{\phi}}, \\ & R_{\hat{\phi}}^{\hat{\theta}}=\boldsymbol{d} \boldsymbol{\omega}_{\hat{\phi}}^{\hat{\theta}}+\boldsymbol{\omega}_{\hat{\theta}}^{\hat{\theta}} \wedge \boldsymbol{\omega}_{\hat{\boldsymbol{\phi}}}^{\hat{\phi}}, \\ & \mathcal{R}_{\hat{r}}^{\hat{\phi}}=\boldsymbol{d} \boldsymbol{\omega}_{\hat{r}}^{\hat{\phi}}+\boldsymbol{\omega}_{\hat{\theta}}^{\hat{\phi}} \wedge \boldsymbol{\omega}_{\hat{r}}^{\hat{\theta}}, \\ & \mathcal{R}_{\hat{\theta}}^{\hat{r}}=\boldsymbol{d} \boldsymbol{\omega}_{\hat{\theta}}^{\hat{r}}+\boldsymbol{\omega}_{\hat{\phi}}^{\hat{r}} \wedge \boldsymbol{\omega}_{\hat{\theta}}^{\hat{\phi}} . \end{aligned}Rr^r^=dωr^,Rθ^i^=ωr^t^ωθ^^^^Rtϕ=ωt^r^ωr^ϕ^,Rϕ^θ^=dωϕ^θ^+ωθ^θ^ωϕ^ϕ^,Rr^ϕ^=dωr^ϕ^+ωθ^ϕ^ωr^θ^,Rθ^r^=dωθ^r^+ωϕ^r^ωθ^ϕ^.
17 We 17 We ^(17)We{ }^{17} \mathrm{We}17We can therefore say R k i k ^ t ^ = R k i k ^ t ^ = R^(k)_(i hat(k) hat(t))=R^{k}{ }_{i \hat{k} \hat{t}}=Rkik^t^= R k ^ t ^ k ^ t ^ = R t ^ k ^ k ^ k ^ = R t ^ k ^ k ^ k ^ R k ^ t ^ k ^ t ^ = R t ^ k ^ k ^ k ^ = R t ^ k ^ k ^ k ^ R_( hat(k) hat(t) hat(k) hat(t))=R_( hat(t) hat(k) hat(k) hat(k))=-R^( hat(t) hat(k) hat(k) hat(k))R_{\hat{k} \hat{t} \hat{k} \hat{t}}=R_{\hat{t} \hat{k} \hat{k} \hat{k}}=-R^{\hat{t} \hat{k} \hat{k} \hat{k}}Rk^t^k^t^=Rt^k^k^k^=Rt^k^k^k^.
18 18 ^(18){ }^{18}18 We write the expressions in this form for simplicity, where the index is raised for simplicity, where the index is raised by the appropriate components of the
Minkowski tensor, since we're working Minkowski tensor, since w
in the orthonormal frame.
in the orthonormal frame.
9 9 ^(9){ }^{9}9 Apply the rule
R μ ^ ν ^ = R | α ^ β ^ | μ ^ ω ^ ω ¯ α ω ~ β R μ ^ ν ^ = R | α ^ β ^ | μ ^ ω ^ ω ¯ α ω ~ β R^( hat(mu) hat(nu))=R_(| hat(alpha) hat(beta)|)^( hat(mu) hat(omega)) bar(omega)^(alpha)^^ tilde(omega)^(beta)\mathcal{R}^{\hat{\mu} \hat{\nu}}=R_{|\hat{\alpha} \hat{\beta}|}^{\hat{\mu} \hat{\omega}} \bar{\omega}^{\alpha} \wedge \tilde{\boldsymbol{\omega}}^{\beta}Rμ^ν^=R|α^β^|μ^ω^ω¯αω~β
The only non-trivial component is
R ϕ ^ r ^ = F ¯ ω ~ r ^ ω ~ ϕ ^ = R r ^ ϕ ^ ϕ ^ ω ^ r ~ r ^ ω ~ ϕ ^ , R ϕ ^ r ^ = F ¯ ω ~ r ^ ω ~ ϕ ^ = R r ^ ϕ ^ ϕ ^ ω ^ r ~ r ^ ω ~ ϕ ^ , R^( hat(phi) hat(r))=- bar(F) tilde(omega)^( hat(r))^^ tilde(omega)^( hat(phi))=R_( hat(r) hat(phi) hat(phi) hat(omega)) tilde(r)^( hat(r))^^ tilde(omega)^( hat(phi)),\mathcal{R}^{\hat{\phi} \hat{r}}=-\bar{F} \tilde{\boldsymbol{\omega}}^{\hat{r}} \wedge \tilde{\boldsymbol{\omega}}^{\hat{\phi}}=R_{\hat{\boldsymbol{r}} \hat{\phi} \hat{\boldsymbol{\phi}} \hat{\boldsymbol{\omega}}} \tilde{\boldsymbol{r}}^{\hat{r}} \wedge \tilde{\boldsymbol{\omega}}^{\hat{\phi}},Rϕ^r^=F¯ω~r^ω~ϕ^=Rr^ϕ^ϕ^ω^r~r^ω~ϕ^,
where we are forced to rearrange the components in order to obey the instruction | α ^ β ^ | | α ^ β ^ | | hat(alpha) hat(beta)||\hat{\alpha} \hat{\beta}||α^β^|
20 20 ^(20){ }^{20}20 The simplest way to do this is to use the rule from Chapter 21 that
G 0 0 = ( R 12 12 + R 23 23 + R 31 31 ) , G 1 1 = ( R 02 02 + R 03 03 + R 23 23 ) , G 0 1 = R 02 12 + R 03 13 , G 1 2 = R 10 20 + R 13 23 , G 0 0 = R 12 12 + R 23 23 + R 31 31 , G 1 1 = R 02 02 + R 03 03 + R 23 23 , G 0 1 = R 02 12 + R 03 13 , G 1 2 = R 10 20 + R 13 23 , {:[G^(0)_(0)=-(R^(12)_(12)+R^(23)_(23)+R^(31)_(31))","],[G^(1)_(1)=-(R^(02)_(02)+R^(03)_(03)+R^(23)_(23))","],[G^(0)_(1)=R^(02)_(12)+R^(03)_(13)","],[G^(1)_(2)=R^(10)_(20)+R^(13)_(23)","]:}\begin{aligned} & G^{0}{ }_{0}=-\left(R^{12}{ }_{12}+R^{23}{ }_{23}+R^{31}{ }_{31}\right), \\ & G^{1}{ }_{1}=-\left(R^{02}{ }_{02}+R^{03}{ }_{03}+R^{23}{ }_{23}\right), \\ & G^{0}{ }_{1}=R^{02}{ }_{12}+R^{03}{ }_{13}, \\ & G^{1}{ }_{2}=R^{10}{ }_{20}+R^{13}{ }_{23}, \end{aligned}G00=(R1212+R2323+R3131),G11=(R0202+R0303+R2323),G01=R0212+R0313,G12=R1020+R1323,
where other components can be found using cyclic permutations.
We find 18 18 ^(18){ }^{18}18
(36.87) R t ^ r ^ = E ω t ^ ω r ^ , R t ^ θ ^ = E ¯ ω t ^ ω θ ^ , R t ^ ϕ ^ = E ¯ ω t ^ ω ϕ ^ , R θ ^ ϕ ^ = F ω θ ^ ω ϕ ^ , R ϕ ^ r ^ = F ¯ ω ϕ ^ ω r ^ , R r ^ θ ^ = F ¯ ω r ^ ω θ ^ , (36.87) R t ^ r ^ = E ω t ^ ω r ^ , R t ^ θ ^ = E ¯ ω t ^ ω θ ^ , R t ^ ϕ ^ = E ¯ ω t ^ ω ϕ ^ , R θ ^ ϕ ^ = F ω θ ^ ω ϕ ^ , R ϕ ^ r ^ = F ¯ ω ϕ ^ ω r ^ , R r ^ θ ^ = F ¯ ω r ^ ω θ ^ , {:(36.87){:[R^( hat(t) hat(r))=Eomega^( hat(t))^^omega^( hat(r))",",R^( hat(t) hat(theta))= bar(E)omega^( hat(t))^^omega^( hat(theta))",",R^( hat(t) hat(phi))= bar(E)omega^( hat(t))^^omega^( hat(phi))","],[R^( hat(theta) hat(phi))=Fomega^( hat(theta))^^omega^( hat(phi))",",R^( hat(phi) hat(r))= bar(F)omega^( hat(phi))^^omega^( hat(r))",",R^( hat(r) hat(theta))= bar(F)omega^( hat(r))^^omega^( hat(theta))","]:}:}\begin{array}{lll} \mathcal{R}^{\hat{t} \hat{r}}=E \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{r}}, & \mathcal{R}^{\hat{t} \hat{\theta}}=\bar{E} \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{\theta}}, & \mathcal{R}^{\hat{t} \hat{\phi}}=\bar{E} \boldsymbol{\omega}^{\hat{t}} \wedge \boldsymbol{\omega}^{\hat{\phi}}, \tag{36.87}\\ \mathcal{R}^{\hat{\theta} \hat{\phi}}=F \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\phi}}, & \mathcal{R}^{\hat{\phi} \hat{r}}=\bar{F} \boldsymbol{\omega}^{\hat{\phi}} \wedge \boldsymbol{\omega}^{\hat{r}}, & \mathcal{R}^{\hat{r} \hat{\theta}}=\bar{F} \boldsymbol{\omega}^{\hat{r}} \wedge \boldsymbol{\omega}^{\hat{\theta}}, \end{array}(36.87)Rt^r^=Eωt^ωr^,Rt^θ^=E¯ωt^ωθ^,Rt^ϕ^=E¯ωt^ωϕ^,Rθ^ϕ^=Fωθ^ωϕ^,Rϕ^r^=F¯ωϕ^ωr^,Rr^θ^=F¯ωr^ωθ^,
where
E = e 2 Λ ( Φ + Φ 2 Φ Λ ) , E ¯ = 1 r e 2 Λ Φ , F = 1 r 2 ( 1 e 2 Λ ) , (36.88) F ¯ = 1 r e 2 Λ Λ . E = e 2 Λ Φ + Φ 2 Φ Λ , E ¯ = 1 r e 2 Λ Φ , F = 1 r 2 1 e 2 Λ , (36.88) F ¯ = 1 r e 2 Λ Λ . {:[E=-e^(-2Lambda)(Phi^('')+Phi^('2)-Phi^(')Lambda^('))","],[ bar(E)=-(1)/(r)e^(-2Lambda)Phi^(')","],[F=(1)/(r^(2))(1-e^(-2Lambda))","],[(36.88) bar(F)=(1)/(r)e^(-2Lambda)Lambda^(').]:}\begin{align*} E & =-\mathrm{e}^{-2 \Lambda}\left(\Phi^{\prime \prime}+\Phi^{\prime 2}-\Phi^{\prime} \Lambda^{\prime}\right), \\ \bar{E} & =-\frac{1}{r} \mathrm{e}^{-2 \Lambda} \Phi^{\prime}, \\ F & =\frac{1}{r^{2}}\left(1-\mathrm{e}^{-2 \Lambda}\right), \\ \bar{F} & =\frac{1}{r} \mathrm{e}^{-2 \Lambda} \Lambda^{\prime} . \tag{36.88} \end{align*}E=e2Λ(Φ+Φ2ΦΛ),E¯=1re2ΛΦ,F=1r2(1e2Λ),(36.88)F¯=1re2ΛΛ.
We can extract the components of the Riemann tensor 19 19 ^(19){ }^{19}19
(36.89) R i r ^ i r ^ = E , R t ^ θ ^ θ ^ i θ ^ = E ¯ , R t ϕ ^ t ϕ ^ = E ¯ , R θ ^ ϕ ^ θ ^ ϕ ^ = F , R r ^ ϕ ^ ϕ ^ ϕ ^ r ^ = F ¯ , R r ^ θ ^ θ ^ r ^ F ^ = F ¯ . (36.89) R i r ^ i r ^ = E , R t ^ θ ^ θ ^ i θ ^ = E ¯ , R t ϕ ^ t ϕ ^ = E ¯ , R θ ^ ϕ ^ θ ^ ϕ ^ = F , R r ^ ϕ ^ ϕ ^ ϕ ^ r ^ = F ¯ , R r ^ θ ^ θ ^ r ^ F ^ = F ¯ . {:(36.89){:[R^(i hat(r))_(i)^( hat(r))=E",",R_( hat(t) hat(theta) hat(theta))^(i hat(theta))= bar(E)",",R^(t hat(phi))_(t hat(phi))= bar(E)","],[R_( hat(theta) hat(phi))^( hat(theta) hat(phi))=F",",R_( hat(r) hat(phi) hat(phi))^( hat(phi) hat(r))=- bar(F)",",R_( hat(r) hat(theta) hat(theta))^( hat(r) hat(F))= bar(F).]:}:}\begin{array}{lll} R^{i \hat{r}}{ }_{i}^{\hat{r}}=E, & R_{\hat{t} \hat{\theta} \hat{\theta}}^{i \hat{\theta}}=\bar{E}, & R^{t \hat{\phi}}{ }_{t \hat{\phi}}=\bar{E}, \\ R_{\hat{\theta} \hat{\phi}}^{\hat{\theta} \hat{\phi}}=F, & R_{\hat{r} \hat{\phi} \hat{\phi}}^{\hat{\phi} \hat{r}}=-\bar{F}, & R_{\hat{r} \hat{\theta} \hat{\theta}}^{\hat{r} \hat{F}}=\bar{F} . \tag{36.89} \end{array}(36.89)Rir^ir^=E,Rt^θ^θ^iθ^=E¯,Rtϕ^tϕ^=E¯,Rθ^ϕ^θ^ϕ^=F,Rr^ϕ^ϕ^ϕ^r^=F¯,Rr^θ^θ^r^F^=F¯.
This gives us the non-zero components of the Einstein equation 20 20 ^(20){ }^{20}20
G i t = G t ^ t ^ = ( F + 2 F ¯ ) , (36.90) G r ^ r ^ = G r ^ r ^ = ( F + 2 E ¯ ) , G θ ^ θ ^ = G ϕ ^ ϕ ^ = ( E + E ¯ + F ¯ ) G i t = G t ^ t ^ = ( F + 2 F ¯ ) , (36.90) G r ^ r ^ = G r ^ r ^ = ( F + 2 E ¯ ) , G θ ^ θ ^ = G ϕ ^ ϕ ^ = ( E + E ¯ + F ¯ ) {:[G_(i)^(t)=-G_( hat(t) hat(t))],[=-(F+2 bar(F))","],[(36.90)G_( hat(r))^( hat(r))=G_( hat(r) hat(r))],[=-(F+2 bar(E))","],[G_( hat(theta))^( hat(theta))=G_( hat(phi))^( hat(phi))=-(E+ bar(E)+ bar(F))]:}\begin{align*} & G_{i}^{t}=-G_{\hat{t} \hat{t}} \\ &=-(F+2 \bar{F}), \\ & G_{\hat{r}}^{\hat{r}}=G_{\hat{r} \hat{r}} \tag{36.90}\\ &=-(F+2 \bar{E}), \\ & G_{\hat{\theta}}^{\hat{\theta}}=G_{\hat{\phi}}^{\hat{\phi}}=-(E+\bar{E}+\bar{F}) \end{align*}Git=Gt^t^=(F+2F¯),(36.90)Gr^r^=Gr^r^=(F+2E¯),Gθ^θ^=Gϕ^ϕ^=(E+E¯+F¯)
There are several more examples to try in the exercises.
We now have an efficient method to extract curvature of a spacetime from its metric. In the final chapters of this part of the book, we discuss some different aspects of geometric techniques.

Chapter summary

  • To compute the Riemann tensor, work in the orthonormal frame and compute the curvature 1-forms, which have the properties
(36.91) d ω μ + ω μ ν ω ν = 0 , (36.91) d ω μ + ω μ ν ω ν = 0 , {:(36.91)domega^(mu)+omega^(mu)_(nu)^^omega^(nu)=0",":}\begin{equation*} d \omega^{\mu}+\omega^{\mu}{ }_{\nu} \wedge \omega^{\nu}=0, \tag{36.91} \end{equation*}(36.91)dωμ+ωμνων=0,
and ω μ ^ ν ^ = ω ν ^ μ ^ ω μ ^ ν ^ = ω ν ^ μ ^ omega_( hat(mu) hat(nu))=-omega_( hat(nu) hat(mu))\boldsymbol{\omega}_{\hat{\mu} \hat{\nu}}=-\boldsymbol{\omega}_{\hat{\nu} \hat{\mu}}ωμ^ν^=ων^μ^.
  • The curvature 2 -form is given by
(36.92) R ν μ = d ω ν μ + ω α μ ω ν α (36.92) R ν μ = d ω ν μ + ω α μ ω ν α {:(36.92)R_(nu)^(mu)=domega_(nu)^(mu)+omega_(alpha)^(mu)^^omega_(nu)^(alpha):}\begin{equation*} \mathcal{R}_{\nu}^{\mu}=\boldsymbol{d} \boldsymbol{\omega}_{\nu}^{\mu}+\boldsymbol{\omega}_{\alpha}^{\mu} \wedge \boldsymbol{\omega}_{\nu}^{\alpha} \tag{36.92} \end{equation*}(36.92)Rνμ=dωνμ+ωαμωνα
  • The components of the Riemann tensor are related to the curvature 2-form via
(36.93) R μ ν = R μ ν | α β | ω α ω β . (36.93) R μ ν = R μ ν | α β | ω α ω β . {:(36.93)R^(mu)_(nu)=R^(mu)_(nu|alpha beta|)omega^(alpha)^^omega^(beta).:}\begin{equation*} \mathcal{R}^{\mu}{ }_{\nu}=R^{\mu}{ }_{\nu|\alpha \beta|} \boldsymbol{\omega}^{\alpha} \wedge \boldsymbol{\omega}^{\beta} . \tag{36.93} \end{equation*}(36.93)Rμν=Rμν|αβ|ωαωβ.

Exercises

(36.1) Show that, when working in the orthonormal (36.8) Consider a time-evolving star with a spherically frame,
(a) ω i t = ω i i ω i t = ω i i omega_(i)^(t)=omega_(i)^(i)\boldsymbol{\omega}_{i}^{t}=\boldsymbol{\omega}_{i}^{i}ωit=ωii.
(b) ω j ^ = ω i ^ j ^ ω j ^ = ω i ^ j ^ omega_( hat(j))=-omega_( hat(i))^( hat(j))\boldsymbol{\omega}_{\hat{j}}=-\omega_{\hat{i}}^{\hat{j}}ωj^=ωi^j^.
(36.2) Read off the connection coefficients in an orthonormal frame for flat space described using cylindrical polar coordinates.
(36.3) A surface has a metric with line element
(36.94) d s 2 = d σ 2 + r ( σ ) 2 d ϕ 2 (36.94) d s 2 = d σ 2 + r ( σ ) 2 d ϕ 2 {:(36.94)ds^(2)=dsigma^(2)+r(sigma)^(2)dphi^(2):}\begin{equation*} \boldsymbol{d} \boldsymbol{s}^{2}=\boldsymbol{d} \sigma^{2}+r(\sigma)^{2} \boldsymbol{d} \phi^{2} \tag{36.94} \end{equation*}(36.94)ds2=dσ2+r(σ)2dϕ2
Compute R σ ^ ϕ ^ R σ ^ ϕ ^ R^( hat(sigma))_( hat(phi))\mathcal{R}^{\hat{\sigma}}{ }_{\hat{\phi}}Rσ^ϕ^.
(36.4) The parabolic surface has a metric with line element
(36.95) d s 2 = ( 1 + a 2 r 2 ) d r 2 + r 2 d θ 2 (36.95) d s 2 = 1 + a 2 r 2 d r 2 + r 2 d θ 2 {:(36.95)ds^(2)=(1+a^(2)r^(2))dr^(2)+r^(2)dtheta^(2):}\begin{equation*} \boldsymbol{d} s^{2}=\left(1+a^{2} r^{2}\right) \boldsymbol{d} r^{2}+r^{2} \boldsymbol{d} \theta^{2} \tag{36.95} \end{equation*}(36.95)ds2=(1+a2r2)dr2+r2dθ2
Compute (i) the components of R R R\boldsymbol{R}R, (ii) the components of the Ricci tensor, and (iii) the Ricci scalar.
(36.5) The torus metric has a line element
(36.96) d s 2 = ( c + a cos v ) 2 d u 2 + a 2 d v 2 (36.96) d s 2 = ( c + a cos v ) 2 d u 2 + a 2 d v 2 {:(36.96)ds^(2)=(c+a cos v)^(2)du^(2)+a^(2)dv^(2):}\begin{equation*} \boldsymbol{d} \boldsymbol{s}^{2}=(c+a \cos v)^{2} \boldsymbol{d} u^{2}+a^{2} \boldsymbol{d} v^{2} \tag{36.96} \end{equation*}(36.96)ds2=(c+acosv)2du2+a2dv2
Compute: (a) the components of R R R\boldsymbol{R}R, (b) the components of the Ricci tensor, and (c) the Ricci scalar.
(36.6) The Poincaré half-plane has line element
(36.97) d s 2 = d x 2 r 2 + d r 2 r 2 (36.97) d s 2 = d x 2 r 2 + d r 2 r 2 {:(36.97)ds^(2)=(dx^(2))/(r^(2))+(dr^(2))/(r^(2)):}\begin{equation*} \boldsymbol{d} s^{2}=\frac{\boldsymbol{d} x^{2}}{r^{2}}+\frac{\boldsymbol{d} r^{2}}{r^{2}} \tag{36.97} \end{equation*}(36.97)ds2=dx2r2+dr2r2
Compute (a) the components of R R R\boldsymbol{R}R, (b) the components of the Ricci tensor, and (c) the Ricci scalar.
(36.7) Consider the expanding universe of the Friedman model, with metric line element
d s 2 = d t 2 + a ( t ) 2 [ d χ 2 + sin 2 χ ( d θ 2 + sin 2 θ d ϕ 2 ) ] d s 2 = d t 2 + a ( t ) 2 d χ 2 + sin 2 χ d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-dt^(2)+a(t)^(2)[dchi^(2)+sin^(2)chi(dtheta^(2)+sin^(2)theta dphi^(2))]\boldsymbol{d} s^{2}=-\boldsymbol{d} t^{2}+a(t)^{2}\left[\boldsymbol{d} \chi^{2}+\sin ^{2} \chi\left(\boldsymbol{d} \theta^{2}+\sin ^{2} \theta \boldsymbol{d} \phi^{2}\right)\right]ds2=dt2+a(t)2[dχ2+sin2χ(dθ2+sin2θdϕ2)].
(36.98)
Compute (a) R ν ^ ν ~ R ν ^ ν ~ R_( hat(nu))^( tilde(nu))\mathcal{R}_{\hat{\nu}}^{\tilde{\nu}}Rν^ν~, (b) the Ricci scalar and, (c) the components of the Einstein tensor G G G\boldsymbol{G}G in the orthonormal frame.
symmetric, time-dependent metric line element
d s 2 = e 2 Φ d T 2 + e 2 Λ d R 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) d s 2 = e 2 Φ d T 2 + e 2 Λ d R 2 + r 2 d θ 2 + sin 2 θ d ϕ 2 ds^(2)=-e^(2Phi)dT^(2)+e^(2Lambda)dR^(2)+r^(2)(dtheta^(2)+sin^(2)theta dphi^(2))\boldsymbol{d} \boldsymbol{s}^{2}=-\mathrm{e}^{2 \Phi} \boldsymbol{d} T^{2}+\mathrm{e}^{2 \Lambda} \boldsymbol{d} R^{2}+r^{2}\left(\boldsymbol{d} \theta^{2}+\sin ^{2} \theta \boldsymbol{d} \phi^{2}\right)ds2=e2ΦdT2+e2ΛdR2+r2(dθ2+sin2θdϕ2), (36.99)
where Φ ( R , T ) , Λ ( R , T ) Φ ( R , T ) , Λ ( R , T ) Phi(R,T),Lambda(R,T)\Phi(R, T), \Lambda(R, T)Φ(R,T),Λ(R,T) and r ( R , T ) r ( R , T ) r(R,T)r(R, T)r(R,T) are all functions of the coordinates R R RRR and T T TTT.
(a) Show that
(36.100) ω T ^ R ^ = Φ e Φ e Λ d T + Λ ˙ e Λ e Φ d R (36.100) ω T ^ R ^ = Φ e Φ e Λ d T + Λ ˙ e Λ e Φ d R {:(36.100)omega_( hat(T))^( hat(R))=Phi^(')e^(Phi)e^(-Lambda)dT+Lambda^(˙)e^(Lambda)e^(-Phi)dR:}\begin{equation*} \boldsymbol{\omega}_{\hat{T}}^{\hat{R}}=\Phi^{\prime} \mathrm{e}^{\Phi} \mathrm{e}^{-\Lambda} \boldsymbol{d} T+\dot{\Lambda} \mathrm{e}^{\Lambda} \mathrm{e}^{-\Phi} \boldsymbol{d} R \tag{36.100} \end{equation*}(36.100)ωT^R^=ΦeΦeΛdT+Λ˙eΛeΦdR
where dashes denote derivatives with respect to R R RRR and dots denote derivatives with respect to T T TTT.
(b) Show that the curvature 2-forms are given by
R R ^ T ^ = E ω T ^ ω R ^ R θ ^ T ^ = E ¯ ω T ^ ω θ ^ + H ω R ^ ω θ ^ R ϕ ^ T ^ = E ¯ ω T ^ ω ϕ ^ + H ω R ^ ω ϕ ^ R ϕ ^ θ ^ = F ω θ ^ ω ϕ ^ R θ ^ θ ^ = F ¯ ω R ^ ω θ ^ H ω T ^ ω θ ^ R ϕ ^ R ^ = F ¯ ω R ^ ω ϕ ^ H ω T ^ ω ϕ ^ R R ^ T ^ = E ω T ^ ω R ^ R θ ^ T ^ = E ¯ ω T ^ ω θ ^ + H ω R ^ ω θ ^ R ϕ ^ T ^ = E ¯ ω T ^ ω ϕ ^ + H ω R ^ ω ϕ ^ R ϕ ^ θ ^ = F ω θ ^ ω ϕ ^ R θ ^ θ ^ = F ¯ ω R ^ ω θ ^ H ω T ^ ω θ ^ R ϕ ^ R ^ = F ¯ ω R ^ ω ϕ ^ H ω T ^ ω ϕ ^ {:[R_( hat(R))^( hat(T))=Eomega^( hat(T))^^omega^( hat(R))],[R_( hat(theta))^( hat(T))= bar(E)omega^( hat(T))^^omega^( hat(theta))+Homega^( hat(R))^^omega^( hat(theta))],[R_( hat(phi))^( hat(T))= bar(E)omega^( hat(T))^^omega^( hat(phi))+Homega^( hat(R))^^omega^( hat(phi))],[R_( hat(phi))^( hat(theta))=Fomega^( hat(theta))^^omega^( hat(phi))],[R_( hat(theta))^( hat(theta))= bar(F)omega^( hat(R))^^omega^( hat(theta))-Homega^( hat(T))^^omega^( hat(theta))],[R_( hat(phi))^( hat(R))= bar(F)omega^( hat(R))^^omega^( hat(phi))-Homega^( hat(T))^^omega^( hat(phi))]:}\begin{aligned} & \mathcal{R}_{\hat{R}}^{\hat{T}}=E \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{R}} \\ & \mathcal{R}_{\hat{\theta}}^{\hat{T}}=\bar{E} \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{\theta}}+H \boldsymbol{\omega}^{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \\ & \mathcal{R}_{\hat{\phi}}^{\hat{T}}=\bar{E} \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{\phi}}+H \boldsymbol{\omega}^{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \\ & \mathcal{R}_{\hat{\phi}}^{\hat{\theta}}=F \boldsymbol{\omega}^{\hat{\theta}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \\ & \mathcal{R}_{\hat{\theta}}^{\hat{\theta}}=\bar{F} \boldsymbol{\omega}^{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{\theta}}-H \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{\theta}} \\ & \mathcal{R}_{\hat{\phi}}^{\hat{R}}=\bar{F} \boldsymbol{\omega}^{\hat{R}} \wedge \boldsymbol{\omega}^{\hat{\phi}}-H \boldsymbol{\omega}^{\hat{T}} \wedge \boldsymbol{\omega}^{\hat{\phi}} \end{aligned}RR^T^=EωT^ωR^Rθ^T^=E¯ωT^ωθ^+HωR^ωθ^Rϕ^T^=E¯ωT^ωϕ^+HωR^ωϕ^Rϕ^θ^=Fωθ^ωϕ^Rθ^θ^=F¯ωR^ωθ^HωT^ωθ^Rϕ^R^=F¯ωR^ωϕ^HωT^ωϕ^
where
E = e 2 Φ ( Λ ¨ + Λ ˙ 2 Λ ˙ Φ ˙ ) e 2 Λ ( Φ + Φ 2 Φ Λ ) E = e 2 Φ Λ ¨ + Λ ˙ 2 Λ ˙ Φ ˙ e 2 Λ Φ + Φ 2 Φ Λ E=e^(-2Phi)((Lambda^(¨))+Lambda^(˙)^(2)-(Lambda^(˙))(Phi^(˙)))-e^(-2Lambda)(Phi^('')+Phi^('2)-Phi^(')Lambda^('))E=\mathrm{e}^{-2 \Phi}\left(\ddot{\Lambda}+\dot{\Lambda}^{2}-\dot{\Lambda} \dot{\Phi}\right)-\mathrm{e}^{-2 \Lambda}\left(\Phi^{\prime \prime}+\Phi^{\prime 2}-\Phi^{\prime} \Lambda^{\prime}\right)E=e2Φ(Λ¨+Λ˙2Λ˙Φ˙)e2Λ(Φ+Φ2ΦΛ),
E ¯ = 1 r e 2 Φ ( r ¨ r ˙ Φ ˙ ) 1 r e 2 Λ r Φ E ¯ = 1 r e 2 Φ ( r ¨ r ˙ Φ ˙ ) 1 r e 2 Λ r Φ bar(E)=(1)/(r)e^(-2Phi)(r^(¨)-r^(˙)Phi^(˙))-(1)/(r)e^(-2Lambda)r^(')Phi^(')\bar{E}=\frac{1}{r} \mathrm{e}^{-2 \Phi}(\ddot{r}-\dot{r} \dot{\Phi})-\frac{1}{r} \mathrm{e}^{-2 \Lambda} r^{\prime} \Phi^{\prime}E¯=1re2Φ(r¨r˙Φ˙)1re2ΛrΦ,
H = 1 r e Φ e Λ ( r ˙ r ˙ Φ r Λ ˙ ) H = 1 r e Φ e Λ r ˙ r ˙ Φ r Λ ˙ H=(1)/(r)e^(-Phi)e^(-Lambda)(r^(˙)^(')-(r^(˙))Phi^(')-r^(')(Lambda^(˙)))H=\frac{1}{r} \mathrm{e}^{-\Phi} \mathrm{e}^{-\Lambda}\left(\dot{r}^{\prime}-\dot{r} \Phi^{\prime}-r^{\prime} \dot{\Lambda}\right)H=1reΦeΛ(r˙r˙ΦrΛ˙),
F = 1 r 2 ( 1 ( r ) 2 e 2 Λ + r ˙ 2 e 2 Φ ) F = 1 r 2 1 r 2 e 2 Λ + r ˙ 2 e 2 Φ F=(1)/(r^(2))(1-(r^('))^(2)e^(-2Lambda)+r^(˙)^(2)e^(-2Phi))F=\frac{1}{r^{2}}\left(1-\left(r^{\prime}\right)^{2} \mathrm{e}^{-2 \Lambda}+\dot{r}^{2} \mathrm{e}^{-2 \Phi}\right)F=1r2(1(r)2e2Λ+r˙2e2Φ),
F ¯ = 1 r e 2 Φ r ˙ Λ ˙ + 1 r e 2 Λ ( r Λ r ) F ¯ = 1 r e 2 Φ r ˙ Λ ˙ + 1 r e 2 Λ r Λ r bar(F)=(1)/(r)e^(-2Phi)r^(˙)Lambda^(˙)+(1)/(r)e^(-2Lambda)(r^(')Lambda^(')-r^(''))\bar{F}=\frac{1}{r} \mathrm{e}^{-2 \Phi} \dot{r} \dot{\Lambda}+\frac{1}{r} \mathrm{e}^{-2 \Lambda}\left(r^{\prime} \Lambda^{\prime}-r^{\prime \prime}\right)F¯=1re2Φr˙Λ˙+1re2Λ(rΛr).
See Misner, Thorne, and Wheeler Chapters 14, 26, and 32 for further details, including the use of this metric in studying stellar pulsations and gravitational collapse.

37

37.1 Motivation: 2-forms and flux
37.2 Hodge star operation 387 37.3 Volume forms 392 Chapter summary 395
Exercises 395
\curvearrowright The material in this chapter is useful in understanding the geometrical interpretation of electromagnetism (Chapter 42) and the Bianchi identity (Chapter 43).
1 1 ^(1){ }^{1}1 In Chapter 32 we saw that an antisymmetric tensor, formed from N N NNN vec tors wedged together is known as a N vector. It is an antisymmetric ( N , 0 N , 0 N,0N, 0N,0 ) tensor.
2 2 ^(2){ }^{2}2 Sir William V. D. Hodge (19031975). The influential mathematician 1975). The influential mathematician
Sir Michael Atiyah was one of his docSir Michael Atiyah was one of his doc-
toral students. Hodge's three-volume toral students. Hodge's three-volume
work Methods of Algebraic Geometry, cowritten with Daniel Pedoe (19101998), freely used the component notation referred to by Cartan in the quotation above.

Duality and the volume form

tensors that encode the same information. By this we mean that we start with one sort of tensor and uniquely obtain another that expresses related physical content. In this section, we shall see how to map between a p p ppp-form, which is an ( 0 , p ) ( 0 , p ) (0,p)(0, p)(0,p) antisymmetric tensor that can built from 1 -forms using the wedge product, and an antisymmetric q q qqq-vector ( q , 0 ) ( q , 0 ) (q,0)(q, 0)(q,0), built from vectors, again combined using the wedge product. 1 1 ^(1){ }^{1}1 This mapping is known as a duality, and can be carried out using an operation known as the Hodge star. 2 2 ^(2){ }^{2}2 Our rather formal discussion in this chapter provides a general method of mapping between two different sorts of tensor. The payoff from this formalism will be a new insight into how volumes can be encoded in tensor algebra.

37.1 Motivation: 2-forms and flux

Let's consider a 2 -form in three-dimensional Euclidean space with components
(37.1) F ~ ( , ) = F 1 ( d y d z ) + F 2 ( d z d x ) + F 3 ( d x d y ) (37.1) F ~ ( , ) = F 1 ( d y d z ) + F 2 ( d z d x ) + F 3 ( d x d y ) {:(37.1) tilde(F)(",")=F^(1)(dy^^dz)+F^(2)(dz^^dx)+F^(3)(dx^^dy):}\begin{equation*} \tilde{\boldsymbol{F}}(,)=F^{1}(\boldsymbol{d} y \wedge \boldsymbol{d} z)+F^{2}(\boldsymbol{d} z \wedge \boldsymbol{d} x)+F^{3}(\boldsymbol{d} x \wedge \boldsymbol{d} y) \tag{37.1} \end{equation*}(37.1)F~(,)=F1(dydz)+F2(dzdx)+F3(dxdy)
By virtue of working in space with n = 3 n = 3 n=3n=3n=3 dimensions, this object has three components. Now let's extract the components and use them to form a vector u u u\boldsymbol{u}u, which we shall write
(37.2) u ( ) = F 1 e x + F 2 e y + F 3 e z (37.2) u ( ) = F 1 e x + F 2 e y + F 3 e z {:(37.2)u()=F^(1)e_(x)+F^(2)e_(y)+F^(3)e_(z):}\begin{equation*} \boldsymbol{u}()=F^{1} \boldsymbol{e}_{x}+F^{2} \boldsymbol{e}_{y}+F^{3} \boldsymbol{e}_{z} \tag{37.2} \end{equation*}(37.2)u()=F1ex+F2ey+F3ez
What is the relationship between F ~ F ~ tilde(F)\tilde{\boldsymbol{F}}F~ and u u u\boldsymbol{u}u ? We shall see shortly that these objects are dual to each other and can be related via the Hodge star operation, written as F ~ = u F ~ = u *** tilde(F)=u\star \tilde{\boldsymbol{F}}=\boldsymbol{u}F~=u. Physically, we can see how they are related by using the notion of flux.
We shall consider a fluid flowing throughout a volume with velocity u u u\boldsymbol{u}u. The flux of the fluid is the amount of fluid flowing through an area
A A AAA in unit time. If the area is a parallelogram with sides given by vectors a a a\boldsymbol{a}a and b b b\boldsymbol{b}b, then shall show in Example 37.1 that
(37.3) ( Flux of u through A ) = F ~ ( a , b ) . (37.3) (  Flux of  u  through  A ) = F ~ ( a , b ) {:(37.3)(" Flux of "u" through "A)= tilde(F)(a","b)". ":}\begin{equation*} (\text { Flux of } \boldsymbol{u} \text { through } A)=\tilde{\boldsymbol{F}}(\boldsymbol{a}, \boldsymbol{b}) \text {. } \tag{37.3} \end{equation*}(37.3)( Flux of u through A)=F~(a,b)
From our discussion in Example 32.6 of Chapter 32, it follows that we can determine the flux by computing F ~ ( a , b ) = F ~ , a b F ~ ( a , b ) = F ~ , a b tilde(F)(a,b)=(: tilde(F),a^^b:)\tilde{\boldsymbol{F}}(\boldsymbol{a}, \boldsymbol{b})=\langle\tilde{\boldsymbol{F}}, \boldsymbol{a} \wedge \boldsymbol{b}\rangleF~(a,b)=F~,ab, that is, the inner product of the 2 -form F F F\boldsymbol{F}F and the bivector a b a b a^^b\boldsymbol{a} \wedge \boldsymbol{b}ab. However, in the next example, we show that the same result can be found using two related objects: the vector u u u\boldsymbol{u}u constructed from F F F\boldsymbol{F}F and a 1-form S S S\boldsymbol{S}S constructed from the bivector a b a b a^^b\boldsymbol{a} \wedge \boldsymbol{b}ab.

Example 37.1

Suppose the only non-zero component of u u u\boldsymbol{u}u is F 3 F 3 F^(3)F^{3}F3. Then, in three dimensions, the amount of fluid per unit time passing through A A AAA will be given by the projection of u u u\boldsymbol{u}u onto the area: F 3 e z ( a × b ) F 3 e z ( a × b ) F^(3)e_(z)*(a xx b)F^{3} \boldsymbol{e}_{\boldsymbol{z}} \cdot(\boldsymbol{a} \times \boldsymbol{b})F3ez(a×b), where we have used the conventional cross product for three-dimensional space to express the area spanned by the a a a\boldsymbol{a}a and b b b\boldsymbol{b}b. Expanding components we find this is equal to F 3 ( a x b y b x a y ) = F 3 d x ( a ) d y ( b ) F 3 a x b y b x a y = F 3 d x ( a ) d y ( b ) F^(3)(a^(x)b^(y)-b^(x)a^(y))=F^(3)dx(a)^^dy(b)F^{3}\left(a^{x} b^{y}-b^{x} a^{y}\right)=F^{3} \boldsymbol{d} x(\boldsymbol{a}) \wedge \boldsymbol{d} y(\boldsymbol{b})F3(axbybxay)=F3dx(a)dy(b). We now let F 1 F 1 F^(1)F^{1}F1 and F 2 F 2 F^(2)F^{2}F2 take non-zero values and repeat the argument for these components to find that the total flux is given by
F 1 d y ( a ) d z ( b ) + F 2 d z ( a ) d x ( b ) + F 3 d x ( a ) d y ( b ) = F ~ ( a , b ) F 1 d y ( a ) d z ( b ) + F 2 d z ( a ) d x ( b ) + F 3 d x ( a ) d y ( b ) = F ~ ( a , b ) F^(1)dy(a)^^dz(b)+F^(2)dz(a)^^dx(b)+F^(3)dx(a)^^dy(b)= tilde(F)(a,b)F^{1} \boldsymbol{d} y(\boldsymbol{a}) \wedge \boldsymbol{d} z(\boldsymbol{b})+F^{2} \boldsymbol{d} z(\boldsymbol{a}) \wedge \boldsymbol{d} x(\boldsymbol{b})+F^{3} \boldsymbol{d} x(\boldsymbol{a}) \wedge \boldsymbol{d} y(\boldsymbol{b})=\tilde{\boldsymbol{F}}(\boldsymbol{a}, \boldsymbol{b})F1dy(a)dz(b)+F2dz(a)dx(b)+F3dx(a)dy(b)=F~(a,b).
This proves eqn 37.3.
It's useful to note that the bivector a b a b a^^b\boldsymbol{a} \wedge \boldsymbol{b}ab has non-zero components ( a b ) y x = ( a b ) y x = (a^^b)^(yx)=(\boldsymbol{a} \wedge \boldsymbol{b})^{y x}=(ab)yx= a y b z b y a z , ( a b ) z x = a z b x b z a x a y b z b y a z , ( a b ) z x = a z b x b z a x a^(y)b^(z)-b^(y)a^(z),(a^^b)^(zx)=a^(z)b^(x)-b^(z)a^(x)a^{y} b^{z}-b^{y} a^{z},(\boldsymbol{a} \wedge \boldsymbol{b})^{z x}=a^{z} b^{x}-b^{z} a^{x}aybzbyaz,(ab)zx=azbxbzax and ( a b ) x y = a x b y b x a y ( a b ) x y = a x b y b x a y (a^^b)^(xy)=a^(x)b^(y)-b^(x)a^(y)(\boldsymbol{a} \wedge \boldsymbol{b})^{x y}=a^{x} b^{y}-b^{x} a^{y}(ab)xy=axbybxay. We notice that if these were to be arranged to make the components of a 1 -form, we would have
(37.5) S ~ = ( a y b z b y a z ) ω x + ( a z b x b z a x ) ω y + ( a x b y b x a y ) ω z (37.5) S ~ = a y b z b y a z ω x + a z b x b z a x ω y + a x b y b x a y ω z {:(37.5) tilde(S)=(a^(y)b^(z)-b^(y)a^(z))omega^(x)+(a^(z)b^(x)-b^(z)a^(x))omega^(y)+(a^(x)b^(y)-b^(x)a^(y))omega^(z):}\begin{equation*} \tilde{\boldsymbol{S}}=\left(a^{y} b^{z}-b^{y} a^{z}\right) \boldsymbol{\omega}^{x}+\left(a^{z} b^{x}-b^{z} a^{x}\right) \boldsymbol{\omega}^{y}+\left(a^{x} b^{y}-b^{x} a^{y}\right) \boldsymbol{\omega}^{z} \tag{37.5} \end{equation*}(37.5)S~=(aybzbyaz)ωx+(azbxbzax)ωy+(axbybxay)ωz
The flux can then also be expressed as S ~ , u S ~ , u (: tilde(S),u:)\langle\tilde{\boldsymbol{S}}, \boldsymbol{u}\rangleS~,u.
This example demonstrates that, in addition to the 2 -form F ~ F ~ tilde(F)\tilde{\boldsymbol{F}}F~, there is a vector u u u\boldsymbol{u}u that encodes the same information via the physical notion of a flux. The vector u u u\boldsymbol{u}u is the dual of F ~ F ~ tilde(F)\tilde{\boldsymbol{F}}F~. Similarly, the bivector a b a b a^^b\boldsymbol{a} \wedge \boldsymbol{b}ab has a dual 1 -form S ~ S ~ tilde(S)\tilde{\boldsymbol{S}}S~. We can obtain the dual of an object using the Hodge star operation; the object outputted by this operation depends on the dimensionality of the space in which we're working. 3 3 ^(3){ }^{3}3 We often work in n = 3 n = 3 n=3n=3n=3-dimensional space, in which the Hodge star takes a q q qqq-form and outputs a ( 3 q ) ( 3 q ) (3-q)(3-q)(3q)-vector. As a result, the Hodge star of a 2 -form is a vector. Similarly, in three dimensions the Hodge star converts a p p ppp-vector into a ( 3 p ) ( 3 p ) (3-p)(3-p)(3p)-form, so that 4 4 ^(4){ }^{4}4 the bivector is dual to a 1 -form.
In the next section, we describe the procedure required to compute a dual and its components. Although fairly straightforward, this is a slightly tedious business when n 3 n 3 n!=3n \neq 3n3, and so this section can be skipped on a first reading. (Some useful rules are collected in the margin at the end of the next section.)

37.2 Hodge star operation

The duality transformation, also known as the Hodge star operation, is carried out using a special tensor called the volume form. This
3 3 ^(3){ }^{3}3 In general, the Hodge star takes a q q q-q-q form and outputs an ( n q ) ( n q ) (n-q)(n-q)(nq)-vector. It also takes a p p ppp-vector and outputs an ( n p ) ( n p ) (n-p)(n-p)(np) form.
4 4 ^(4){ }^{4}4 This property of three-dimensional Euclidean space is what allows our conEuclidean space is what allows our con-
ventional vector calculus to work as ventional vector calculus to work as
neatly as it does. In particular, it exneatly as it does. In particular, it ex-
plains the existence of the cross product plains the existence of the cross product
a × b a × b a xx b\boldsymbol{a} \times \boldsymbol{b}a×b which is the vector that is dual to a bivector a b a b a^^b\boldsymbol{a} \wedge \boldsymbol{b}ab.
5 5 ^(5){ }^{5}5 Recall that the orthonormal basis is an example of a local frame in which measurements can be made. For (3+1)dimensional spacetime it has metric tensor
g = ( ω 0 ^ ) ( ω 0 ^ ) + ( ω 1 ^ ) ( ω 1 ^ ) g = ω 0 ^ ω 0 ^ + ω 1 ^ ω 1 ^ g=-(omega^( hat(0)))ox(omega^( hat(0)))+(omega^( hat(1)))ox(omega^( hat(1)))\boldsymbol{g}=-\left(\boldsymbol{\omega}^{\hat{0}}\right) \otimes\left(\boldsymbol{\omega}^{\hat{0}}\right)+\left(\boldsymbol{\omega}^{\hat{1}}\right) \otimes\left(\boldsymbol{\omega}^{\hat{1}}\right)g=(ω0^)(ω0^)+(ω1^)(ω1^)
+ ( ω 2 ^ ) ( ω 2 ^ ) + ( ω 3 ) ( ω 3 ) + ω 2 ^ ω 2 ^ + ω 3 ω 3 +(omega^( hat(2)))ox(omega^( hat(2)))+(omega^(3))ox(omega^(3))+\left(\boldsymbol{\omega}^{\hat{2}}\right) \otimes\left(\boldsymbol{\omega}^{\hat{2}}\right)+\left(\boldsymbol{\omega}^{3}\right) \otimes\left(\boldsymbol{\omega}^{3}\right)+(ω2^)(ω2^)+(ω3)(ω3)
which has signature ( + + + ) ( + + + ) (-+++)(-+++)(+++).
6 6 ^(6){ }^{6}6 It's a candidate at this stage as we don't yet have a means of determining the ordering of the basis 1 -forms. This gives rise to a sign ambiguity that we resolve in the next Example.
is another example of a p p ppp-form, this one built from the basis forms of the space in which we're working. Working in the four-dimensional orthonormal basis 5 5 ^(5){ }^{5}5 with basis 1 -forms ω 0 ^ , , ω 3 ^ ω 0 ^ , , ω 3 ^ omega^( hat(0)),dots,omega^( hat(3))\boldsymbol{\omega}^{\hat{0}}, \ldots, \boldsymbol{\omega}^{\hat{3}}ω0^,,ω3^. We then construct a candidate volume 4 -form ω ~ ω ~ tilde(omega)\tilde{\boldsymbol{\omega}}ω~ via 6 6 ^(6)^{6}6
(37.6) ω ~ = ω μ ^ ω ν ^ ω α ^ ω β ^ . (37.6) ω ~ = ω μ ^ ω ν ^ ω α ^ ω β ^ . {:(37.6) tilde(omega)=omega^( hat(mu))^^omega^( hat(nu))^^omega^( hat(alpha))^^omega^( hat(beta)).:}\begin{equation*} \tilde{\omega}=\omega^{\hat{\mu}} \wedge \omega^{\hat{\nu}} \wedge \omega^{\hat{\alpha}} \wedge \omega^{\hat{\beta}} . \tag{37.6} \end{equation*}(37.6)ω~=ωμ^ων^ωα^ωβ^.
In the orthonormal basis, the volume form has components ω μ ^ ν ^ α ^ β ^ = ω μ ^ ν ^ α ^ β ^ = omega_( hat(mu) hat(nu) hat(alpha) hat(beta))=\omega_{\hat{\mu} \hat{\nu} \hat{\alpha} \hat{\beta}}=ωμ^ν^α^β^= ε μ ν α β ε μ ν α β epsi_(mu nu alpha beta)\varepsilon_{\mu \nu \alpha \beta}εμναβ, where ε μ ν α β ε μ ν α β epsi_(mu nu alpha beta)\varepsilon_{\mu \nu \alpha \beta}εμναβ is the four-dimensional Levi-Civita symbol.

Example 37.2

We can treat the Levi-Civita symbol ε μ ν α β ε μ ν α β epsi_(mu nu alpha beta)\varepsilon_{\mu \nu \alpha \beta}εμναβ as being the components of a LeviCivita tensor ε ( , , ε ( , , epsi(,,\varepsilon(,,ε(,,, ) . T h e t e n s o r i s a n t i s y m m e t r i c i n a l l o f i t s s l o t s , s o i t c h a n g e s ) . T h e t e n s o r i s a n t i s y m m e t r i c i n a l l o f i t s s l o t s , s o i t c h a n g e s ).Thetensorisantisymmetricinallofitsslots,soitchanges) . The tensor is antisymmetric in all of its slots, so it changes).Thetensorisantisymmetricinallofitsslots,soitchanges sign when any two vectors exchange their slots. In Minkowski spacetime, we fix the sign by saying
7 7 ^(7){ }^{7}7 We also have the rule, if we have M M MMM indices, that
(37.8) ε i j k ε l m r = M ! δ i j k l m r (37.8) ε i j k ε l m r = M ! δ i j k l m r {:(37.8)epsi_(ij dots k)epsi^(lm dots r)=M!delta_(ij dots k)^(lm dots r):}\begin{equation*} \varepsilon_{i j \ldots k} \varepsilon^{l m \ldots r}=M!\delta_{i j \ldots k}^{l m \ldots r} \tag{37.8} \end{equation*}(37.8)εijkεlmr=M!δijklmr
ε ( e 0 , e 1 , e 2 , e 3 ) = ε 0123 = 1 ε e 0 , e 1 , e 2 , e 3 = ε 0123 = 1 epsi(e_(0),e_(1),e_(2),e_(3))=epsi_(0123)=1\boldsymbol{\varepsilon}\left(e_{0}, \boldsymbol{e}_{1}, \boldsymbol{e}_{2}, \boldsymbol{e}_{3}\right)=\varepsilon_{0123}=1ε(e0,e1,e2,e3)=ε0123=1
The components are then 7 7 ^(7){ }^{7}7
(37.9) ε μ ν α β = { 0 unless μ , ν , α , β are all different + 1 for even permutations of 0 , 1 , 2 , 3 , 1 for odd permutation. (37.9) ε μ ν α β = 0  unless  μ , ν , α , β  are all different  + 1  for even permutations of  0 , 1 , 2 , 3 , 1  for odd permutation.  {:(37.9)epsi_(mu nu alpha beta)={[0" unless "mu","nu","alpha","beta" are all different "],[+1" for even permutations of "0","1","2","3","],[-1" for odd permutation. "]:}:}\varepsilon_{\mu \nu \alpha \beta}=\left\{\begin{array}{l} 0 \text { unless } \mu, \nu, \alpha, \beta \text { are all different } \tag{37.9}\\ +1 \text { for even permutations of } 0,1,2,3, \\ -1 \text { for odd permutation. } \end{array}\right.(37.9)εμναβ={0 unless μ,ν,α,β are all different +1 for even permutations of 0,1,2,3,1 for odd permutation. 
It's important to note that, if we're treating ε μ ν α β ε μ ν α β epsi_(mu nu alpha beta)\varepsilon_{\mu \nu \alpha \beta}εμναβ as the components of a tensor, as we do here, then in Minkowski spacetime the all-up version has the property that
(37.10) ε 0123 = ε 0123 = 1 (37.10) ε 0123 = ε 0123 = 1 {:(37.10)epsi^(0123)=-epsi_(0123)=-1:}\begin{equation*} \varepsilon^{0123}=-\varepsilon_{0123}=-1 \tag{37.10} \end{equation*}(37.10)ε0123=ε0123=1
In terms of components we can then, using the bar notation of the previous chapter, fix the sign by writing the volume 4 -form as
ω ~ = 1 4 ! ε μ ν α β ω μ ^ ω ν ^ ω α ^ ω β ^ = ε | μ ν α β | ω μ ^ ω ν ^ ω α ^ ω β ^ (37.11) = ω 0 ^ ω 1 ^ ω 2 ^ ω 3 ^ ω ~ = 1 4 ! ε μ ν α β ω μ ^ ω ν ^ ω α ^ ω β ^ = ε | μ ν α β | ω μ ^ ω ν ^ ω α ^ ω β ^ (37.11) = ω 0 ^ ω 1 ^ ω 2 ^ ω 3 ^ {:[ tilde(omega)=(1)/(4!)epsi_(mu nu alpha beta)omega^( hat(mu))^^omega^( hat(nu))^^omega^( hat(alpha))^^omega^( hat(beta))=epsi_(|mu nu alpha beta|)omega^( hat(mu))^^omega^( hat(nu))^^omega^( hat(alpha))^^omega^( hat(beta))],[(37.11)=omega^( hat(0))^^omega^( hat(1))^^omega^( hat(2))^^omega^( hat(3))]:}\begin{align*} \tilde{\boldsymbol{\omega}} & =\frac{1}{4!} \varepsilon_{\mu \nu \alpha \beta} \boldsymbol{\omega}^{\hat{\mu}} \wedge \boldsymbol{\omega}^{\hat{\nu}} \wedge \boldsymbol{\omega}^{\hat{\alpha}} \wedge \boldsymbol{\omega}^{\hat{\beta}}=\varepsilon_{|\mu \nu \alpha \beta|} \boldsymbol{\omega}^{\hat{\mu}} \wedge \boldsymbol{\omega}^{\hat{\nu}} \wedge \boldsymbol{\omega}^{\hat{\alpha}} \wedge \boldsymbol{\omega}^{\hat{\beta}} \\ & =\boldsymbol{\omega}^{\hat{0}} \wedge \boldsymbol{\omega}^{\hat{1}} \wedge \boldsymbol{\omega}^{\hat{2}} \wedge \boldsymbol{\omega}^{\hat{3}} \tag{37.11} \end{align*}ω~=14!εμναβωμ^ων^ωα^ωβ^=ε|μναβ|ωμ^ων^ωα^ωβ^(37.11)=ω0^ω1^ω2^ω3^
If, instead of an orthonormal basis, we consider a general metric space, we can use an arbitrary coordinate system with basis 1-forms d x 0 , , d x n d x 0 , , d x n dx^(0),dots,dx^(n)\boldsymbol{d} x^{0}, \ldots, \boldsymbol{d} x^{n}dx0,,dxn. In that case, we must include a factor of the determinant of the metric in our definition of the volume form thus:
(37.12) ω ~ = ( 1 ) s g d x 0 d x 1 d x n (37.12) ω ~ = ( 1 ) s g d x 0 d x 1 d x n {:(37.12) tilde(omega)=sqrt((-1)^(s)g)dx^(0)^^dx^(1)^^dots^^dx^(n):}\begin{equation*} \tilde{\boldsymbol{\omega}}=\sqrt{(-1)^{s} g} \boldsymbol{d} x^{0} \wedge \boldsymbol{d} x^{1} \wedge \ldots \wedge \boldsymbol{d} x^{n} \tag{37.12} \end{equation*}(37.12)ω~=(1)sgdx0dx1dxn
where g g ggg is the determinant of the metric and s s sss is the number of minus signs in the signature of the metric. The ( 1 ) s ( 1 ) s (-1)^(s)(-1)^{s}(1)s factor ensures that we do not attempt to take the square root of a negative number.

Example 37.3

Consider a flat two-dimensional plane expressed in polar coordinates with line element 8 d s 2 = d r 2 + r 2 d θ 2 8 d s 2 = d r 2 + r 2 d θ 2 ^(8)ds^(2)=dr^(2)+r^(2)dtheta^(2){ }^{8} \mathrm{~d} s^{2}=\mathrm{d} r^{2}+r^{2} \mathrm{~d} \theta^{2}8 ds2=dr2+r2 dθ2. Expressed in orthonormal coordinates ( x 1 , x 2 ) = ( r , θ ) x 1 , x 2 = ( r , θ ) (x^(1),x^(2))=(r,theta)\left(x^{1}, x^{2}\right)=(r, \theta)(x1,x2)=(r,θ), we have basis 1 -forms
The volume form is given by
(37.14) ω ~ = ω r ^ ω θ ^ = r d r d θ (37.14) ω ~ = ω r ^ ω θ ^ = r d r d θ {:(37.14) tilde(omega)=omega^( hat(r))^^omega^( hat(theta))=rdr^^d theta:}\begin{equation*} \tilde{\boldsymbol{\omega}}=\boldsymbol{\omega}^{\hat{r}} \wedge \boldsymbol{\omega}^{\hat{\theta}}=r \boldsymbol{d} r \wedge \boldsymbol{d} \theta \tag{37.14} \end{equation*}(37.14)ω~=ωr^ωθ^=rdrdθ
This volume form has components given by
ω 12 = ω ~ ( e r ^ , e θ ^ ) = [ ω r ^ ( e r ^ ) ω θ ^ ( e θ ^ ) ω θ ^ ( e r ^ ) ω r ^ ( e θ ^ ) ] = 1 , (37.15) ω 21 = ω ~ ( e θ ^ , e r ^ ) = [ ω r ^ ( e θ ^ ) ω θ ^ ( e r ^ ) ω θ ^ ( e θ ^ ) ω r ^ ( e r ^ ) ] = 1 , ω 12 = ω ~ e r ^ , e θ ^ = ω r ^ e r ^ ω θ ^ e θ ^ ω θ ^ e r ^ ω r ^ e θ ^ = 1 , (37.15) ω 21 = ω ~ e θ ^ , e r ^ = ω r ^ e θ ^ ω θ ^ e r ^ ω θ ^ e θ ^ ω r ^ e r ^ = 1 , {:[omega_(12)= tilde(omega)(e_( hat(r)),e_( hat(theta)))=[omega^( hat(r))(e_( hat(r)))oxomega^( hat(theta))(e_( hat(theta)))-omega^( hat(theta))(e_( hat(r)))oxomega^( hat(r))(e_( hat(theta)))]=1","],[(37.15)omega_(21)= tilde(omega)(e_( hat(theta)),e_( hat(r)))=[omega^( hat(r))(e_( hat(theta)))oxomega^( hat(theta))(e_( hat(r)))-omega^( hat(theta))(e_( hat(theta)))oxomega^( hat(r))(e_( hat(r)))]=-1","]:}\begin{align*} & \omega_{12}=\tilde{\omega}\left(e_{\hat{r}}, e_{\hat{\theta}}\right)=\left[\omega^{\hat{r}}\left(e_{\hat{r}}\right) \otimes \boldsymbol{\omega}^{\hat{\theta}}\left(e_{\hat{\theta}}\right)-\omega^{\hat{\theta}}\left(e_{\hat{r}}\right) \otimes \boldsymbol{\omega}^{\hat{r}}\left(\boldsymbol{e}_{\hat{\theta}}\right)\right]=1, \\ & \omega_{21}=\tilde{\omega}\left(e_{\hat{\theta}}, e_{\hat{r}}\right)=\left[\omega^{\hat{r}}\left(\boldsymbol{e}_{\hat{\theta}}\right) \otimes \boldsymbol{\omega}^{\hat{\theta}}\left(e_{\hat{r}}\right)-\omega^{\hat{\theta}}\left(\boldsymbol{e}_{\hat{\theta}}\right) \otimes \boldsymbol{\omega}^{\hat{r}}\left(\boldsymbol{e}_{\hat{r}}\right)\right]=-1, \tag{37.15} \end{align*}ω12=ω~(er^,eθ^)=[ωr^(er^)ωθ^(eθ^)ωθ^(er^)ωr^(eθ^)]=1,(37.15)ω21=ω~(eθ^,er^)=[ωr^(eθ^)ωθ^(er^)ωθ^(eθ^)ωr^(er^)]=1,
or
(37.16) ω i j = ε i j . (37.16) ω i j = ε i j . {:(37.16)omega_(ij)=epsi_(ij).:}\begin{equation*} \omega_{i j}=\varepsilon_{i j} . \tag{37.16} \end{equation*}(37.16)ωij=εij.
If, instead, we choose to work with in a coordinate basis with basis 1-forms ω r = d r ω r = d r omega^(r)=dr\boldsymbol{\omega}^{r}=\boldsymbol{d} rωr=dr and ω θ = d θ ω θ = d θ omega^(theta)=d theta\boldsymbol{\omega}^{\theta}=\boldsymbol{d} \thetaωθ=dθ we must multiply d r d θ d r d θ dr^^d theta\boldsymbol{d} r \wedge \boldsymbol{d} \thetadrdθ a factor of g = det g i j = r g = det g i j = r sqrtg=sqrt(detg_(ij))=r\sqrt{g}=\sqrt{\operatorname{det} g_{i j}}=rg=detgij=r and obtain the same result for the volume form ω ~ ω ~ tilde(omega)\tilde{\boldsymbol{\omega}}ω~. Note, however, that in the coordinate basis, the volume form has components g ε i j g ε i j sqrtgepsi_(ij)\sqrt{g} \varepsilon_{i j}gεij, or
(37.17) ω i j = r ε i j (37.17) ω i j = r ε i j {:(37.17)omega_(ij)=repsi_(ij):}\begin{equation*} \omega_{i j}=r \varepsilon_{i j} \tag{37.17} \end{equation*}(37.17)ωij=rεij
With the volume form ω ~ ω ~ tilde(omega)\tilde{\boldsymbol{\omega}}ω~ in hand, we can define the duality transformation using the Hodge star operation. In n n nnn-dimensional space, this operation takes a ( q , 0 ) ( q , 0 ) (q,0)(q, 0)(q,0) tensor and outputs a p p ppp-form, where p = n q p = n q p=n-qp=n-qp=nq. Consider a ( q , 0 ) ( q , 0 ) (q,0)(q, 0)(q,0) antisymmetric tensor T T T\boldsymbol{T}T with components T α κ = T α κ = T^(alpha dots kappa)=T^{\alpha \ldots \kappa}=Tακ= T [ α κ ] T [ α κ ] T^([alpha dots kappa])T^{[\alpha \ldots \kappa]}T[ακ]. The p p ppp-form A ~ A ~ tilde(A)\tilde{\boldsymbol{A}}A~ that is dual to T T T\boldsymbol{T}T is written as
(37.18) A ~ = T = A μ ω p ! ω μ ω ω (37.18) A ~ = T = A μ ω p ! ω μ ω ω {:(37.18) tilde(A)=***T=(A_(mu dots omega))/(p!)omega^(mu)^^dots^^omega^(omega):}\begin{equation*} \tilde{\boldsymbol{A}}=\star \boldsymbol{T}=\frac{A_{\mu \ldots \omega}}{p!} \boldsymbol{\omega}^{\mu} \wedge \ldots \wedge \omega^{\omega} \tag{37.18} \end{equation*}(37.18)A~=T=Aμωp!ωμωω
where the ω μ ω μ omega^(mu)\boldsymbol{\omega}^{\mu}ωμ are the basis 1-forms, p = n q p = n q p=n-qp=n-qp=nq and the dual tensor's components are given by
(37.19) A μ ω = 1 q ! ω α κ μ ω T α κ (37.19) A μ ω = 1 q ! ω α κ μ ω T α κ {:(37.19)A_(mu dots omega)=(1)/(q!)omega_(alpha dots kappa mu dots omega)T^(alpha dots kappa):}\begin{equation*} A_{\mu \ldots \omega}=\frac{1}{q!} \omega_{\alpha \ldots \kappa \mu \ldots \omega} T^{\alpha \ldots \kappa} \tag{37.19} \end{equation*}(37.19)Aμω=1q!ωακμωTακ
Despite the fussy definition, the practice of taking the dual turns out to be rather simple, especially in the usual ( x 0 , , x n ) x 0 , , x n (x^(0),dots,x^(n))\left(x^{0}, \ldots, x^{n}\right)(x0,,xn) Minkowski space. Here the volume form is given by d x 0 d x n d x 0 d x n dx^(0)^^dots^^dx^(n)\boldsymbol{d} x^{0} \wedge \ldots \wedge \boldsymbol{d} x^{n}dx0dxn, and has components given by the Levi-Civita symbol ε 0 n ε 0 n epsi_(0dots n)\varepsilon_{0 \ldots n}ε0n.

Example 37.4

In three-dimensional Euclidean space in Cartesian coordinates, we have the volume 3 -form ω ~ = 1 3 ! ε μ ν σ d x μ d x ν d x σ ω ~ = 1 3 ! ε μ ν σ d x μ d x ν d x σ tilde(omega)=(1)/(3!)epsi_(mu nu sigma)dx^(mu)^^dx^(nu)^^dx^(sigma)\tilde{\boldsymbol{\omega}}=\frac{1}{3!} \varepsilon_{\mu \nu \sigma} \boldsymbol{d} x^{\mu} \wedge \boldsymbol{d} x^{\nu} \wedge \boldsymbol{d} x^{\sigma}ω~=13!εμνσdxμdxνdxσ with components ε μ ν σ ε μ ν σ epsi_(mu nu sigma)\varepsilon_{\mu \nu \sigma}εμνσ, where ε 123 = 1 ε 123 = 1 epsi_(123)=1\varepsilon_{123}=1ε123=1.
Consider a bivector T = e 2 e 3 T = e 2 e 3 T=e_(2)^^e_(3)\boldsymbol{T}=\boldsymbol{e}_{2} \wedge \boldsymbol{e}_{3}T=e2e3, which has components T α β = T ( ω α , ω β ) T α β = T ω α , ω β T^(alpha beta)=T(omega^(alpha),omega^(beta))T^{\alpha \beta}=\boldsymbol{T}\left(\boldsymbol{\omega}^{\alpha}, \boldsymbol{\omega}^{\beta}\right)Tαβ=T(ωα,ωβ) given by
T α β = e 2 ( ω α ) e 3 ( ω β ) e 3 ( ω α ) e 2 ( ω β ) (37.20) = δ 2 α δ 3 β δ 3 α δ 2 β T α β = e 2 ω α e 3 ω β e 3 ω α e 2 ω β (37.20) = δ 2 α δ 3 β δ 3 α δ 2 β {:[T^(alpha beta)=e_(2)(omega^(alpha))oxe_(3)(omega^(beta))-e_(3)(omega^(alpha))oxe_(2)(omega^(beta))],[(37.20)=delta_(2)^(alpha)delta_(3)^(beta)-delta_(3)^(alpha)delta_(2)^(beta)]:}\begin{align*} T^{\alpha \beta} & =\boldsymbol{e}_{2}\left(\boldsymbol{\omega}^{\alpha}\right) \otimes \boldsymbol{e}_{3}\left(\boldsymbol{\omega}^{\beta}\right)-\boldsymbol{e}_{3}\left(\boldsymbol{\omega}^{\alpha}\right) \otimes \boldsymbol{e}_{2}\left(\boldsymbol{\omega}^{\beta}\right) \\ & =\delta_{2}^{\alpha} \delta_{3}^{\beta}-\delta_{3}^{\alpha} \delta_{2}^{\beta} \tag{37.20} \end{align*}Tαβ=e2(ωα)e3(ωβ)e3(ωα)e2(ωβ)(37.20)=δ2αδ3βδ3αδ2β
This q = 2 q = 2 q=2q=2q=2 vector is dual to an object with valence ( 0 , p ) = ( 0 , 3 2 ) = ( 0 , 1 ) ( 0 , p ) = ( 0 , 3 2 ) = ( 0 , 1 ) (0,p)=(0,3-2)=(0,1)(0, p)=(0,3-2)=(0,1)(0,p)=(0,32)=(0,1), that is, a 1 -form A ~ = T A ~ = T tilde(A)=***T\tilde{\boldsymbol{A}}=\star \boldsymbol{T}A~=T. This 1 -form has components
A μ = 1 q ! ε α β μ T α β (37.21) = 1 2 ε α β μ ( δ 2 α δ 3 β δ 3 α δ 2 β ) = 1 2 ( ε 23 μ ε 32 μ ) A μ = 1 q ! ε α β μ T α β (37.21) = 1 2 ε α β μ δ 2 α δ 3 β δ 3 α δ 2 β = 1 2 ε 23 μ ε 32 μ {:[A_(mu)=(1)/(q!)epsi_(alpha beta mu)T^(alpha beta)],[(37.21)=(1)/(2)epsi_(alpha beta mu)(delta_(2)^(alpha)delta_(3)^(beta)-delta_(3)^(alpha)delta_(2)^(beta))=(1)/(2)(epsi_(23 mu)-epsi_(32 mu))]:}\begin{align*} A_{\mu} & =\frac{1}{q!} \varepsilon_{\alpha \beta \mu} T^{\alpha \beta} \\ & =\frac{1}{2} \varepsilon_{\alpha \beta \mu}\left(\delta_{2}^{\alpha} \delta_{3}^{\beta}-\delta_{3}^{\alpha} \delta_{2}^{\beta}\right)=\frac{1}{2}\left(\varepsilon_{23 \mu}-\varepsilon_{32 \mu}\right) \tag{37.21} \end{align*}Aμ=1q!εαβμTαβ(37.21)=12εαβμ(δ2αδ3βδ3αδ2β)=12(ε23με32μ)
9 9 ^(9){ }^{9}9 This fussy sign convention is the one used in most general relativity texts and so we employ it here. It is useful since it allows us to raise the components of ω ω omega\omegaω with the metric as we expect to be able to do with the components of any tensor in a metric space.
We conclude
(37.22) A 1 = 1 2 ( ε 231 ε 321 ) = 1 (37.22) A 1 = 1 2 ε 231 ε 321 = 1 {:(37.22)A_(1)=(1)/(2)(epsi_(231)-epsi_(321))=1:}\begin{equation*} A_{1}=\frac{1}{2}\left(\varepsilon_{231}-\varepsilon_{321}\right)=1 \tag{37.22} \end{equation*}(37.22)A1=12(ε231ε321)=1
with other components vanishing. We have, therefore, that A ~ = ω 1 = d x 1 A ~ = ω 1 = d x 1 tilde(A)=omega^(1)=dx^(1)\tilde{\boldsymbol{A}}=\boldsymbol{\omega}^{1}=\boldsymbol{d} x^{1}A~=ω1=dx1. In brief, ( e y e x ) = d x e y e x = d x ***(e_(y)^^e_(x))=dx\star\left(\boldsymbol{e}_{y} \wedge \boldsymbol{e}_{x}\right)=\boldsymbol{d} x(eyex)=dx in three dimensions.
In order to define the inverse map that inputs the p p ppp-form and outputs a q q qqq-tensor, we need the components ω α κ ω α κ omega^(alpha dots kappa)\omega^{\alpha \ldots \kappa}ωακ of a n n nnn-vector version of the volume n n nnn-form ω ~ ω ~ tilde(omega)\tilde{\boldsymbol{\omega}}ω~. In a general metric space, these components are determined by the equation
(37.23) ω α κ ω α κ = ( 1 ) s n ! (37.23) ω α κ ω α κ = ( 1 ) s n ! {:(37.23)omega^(alpha dots kappa)omega_(alpha dots kappa)=(-1)^(s)n!:}\begin{equation*} \omega^{\alpha \ldots \kappa} \omega_{\alpha \ldots \kappa}=(-1)^{s} n! \tag{37.23} \end{equation*}(37.23)ωακωακ=(1)sn!
where, once again, s s sss is the number of negative signs in the metric's signature. 9 9 ^(9){ }^{9}9 This guarantees the normalization condition that ω 123 n = ω 123 n = omega^(123 dots n)=\omega^{123 \ldots n}=ω123n= ( 1 ) s / ω 123 n ( 1 ) s / ω 123 n (-1)^(s)//omega_(123 dots n)(-1)^{s} / \omega_{123 \ldots n}(1)s/ω123n.

Example 37.5

For an orthonormal basis with s = 1 s = 1 s=1s=1s=1 we have that the components of the volume tensor are
(37.24) ω α β κ = ε α β κ (37.24) ω α β κ = ε α β κ {:(37.24)omega^(alpha beta dots kappa)=epsi^(alpha beta dots kappa):}\begin{equation*} \omega^{\alpha \beta \ldots \kappa}=\varepsilon^{\alpha \beta \ldots \kappa} \tag{37.24} \end{equation*}(37.24)ωαβκ=εαβκ
and so ω 123 n ω 123 n = ε α β κ ε α β κ = n ω 123 n ω 123 n = ε α β κ ε α β κ = n omega^(123 dots n)omega_(123 dots n)=epsi^(alpha beta dots kappa)epsi_(alpha beta dots kappa)=-n\omega^{123 \ldots n} \omega_{123 \ldots n}=\varepsilon^{\alpha \beta \ldots \kappa} \varepsilon_{\alpha \beta \ldots \kappa}=-nω123nω123n=εαβκεαβκ=n !
In the usual ( 3 + 1 ) ( 3 + 1 ) (3+1)(3+1)(3+1)-dimensional metric spacetime with s = 1 s = 1 s=1s=1s=1, this gives us the relations
(37.25) ω μ ν α β = g ε μ ν α β , ω μ ν α β = 1 g ε μ ν α β (37.25) ω μ ν α β = g ε μ ν α β , ω μ ν α β = 1 g ε μ ν α β {:(37.25)omega_(mu nu alpha beta)=sqrt(-g)epsi_(mu nu alpha beta)","quadomega^(mu nu alpha beta)=(1)/(sqrt(-g))epsi^(mu nu alpha beta):}\begin{equation*} \omega_{\mu \nu \alpha \beta}=\sqrt{-g} \varepsilon_{\mu \nu \alpha \beta}, \quad \omega^{\mu \nu \alpha \beta}=\frac{1}{\sqrt{-g}} \varepsilon^{\mu \nu \alpha \beta} \tag{37.25} \end{equation*}(37.25)ωμναβ=gεμναβ,ωμναβ=1gεμναβ
where ε 0123 = ε 0123 = 1 ε 0123 = ε 0123 = 1 epsi^(0123)=-epsi_(0123)=-1\varepsilon^{0123}=-\varepsilon_{0123}=-1ε0123=ε0123=1.
In n n nnn-dimensional space, the dual ( q , 0 ) ( q , 0 ) (q,0)(q, 0)(q,0) tensor S S S\boldsymbol{S}S of the p p ppp-form B ~ B ~ tilde(B)\tilde{\boldsymbol{B}}B~ is given by
(37.26) S = B ~ = S α κ q ! e α e κ (37.26) S = B ~ = S α κ q ! e α e κ {:(37.26)S=*** tilde(B)=(S^(alpha dots kappa))/(q!)e_(alpha)^^dots^^e_(kappa):}\begin{equation*} \boldsymbol{S}=\star \tilde{\boldsymbol{B}}=\frac{S^{\alpha \ldots \kappa}}{q!} e_{\alpha} \wedge \ldots \wedge e_{\kappa} \tag{37.26} \end{equation*}(37.26)S=B~=Sακq!eαeκ
where q = n p q = n p q=n-pq=n-pq=np and the components are
(37.27) S α κ = 1 p ! ω μ ω α κ B μ ω (37.27) S α κ = 1 p ! ω μ ω α κ B μ ω {:(37.27)S^(alpha dots kappa)=(1)/(p!)omega^(mu dots omega alpha dots kappa)B_(mu dots omega):}\begin{equation*} S^{\alpha \ldots \kappa}=\frac{1}{p!} \omega^{\mu \ldots \omega \alpha \ldots \kappa} B_{\mu \ldots \omega} \tag{37.27} \end{equation*}(37.27)Sακ=1p!ωμωακBμω

Example 37.6

In flat n = 3 n = 3 n=3n=3n=3-dimensional space, we have the volume 3 -form ω ~ = d x 1 d x 2 d x 3 ω ~ = d x 1 d x 2 d x 3 tilde(omega)=dx^(1)^^dx^(2)^^dx^(3)\tilde{\boldsymbol{\omega}}=\boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3}ω~=dx1dx2dx3, with components ε μ ν σ ε μ ν σ epsi_(mu nu sigma)\varepsilon_{\mu \nu \sigma}εμνσ. The corresponding up version has components ε μ ν σ ε μ ν σ epsi^(mu nu sigma)\varepsilon^{\mu \nu \sigma}εμνσ, with ε 123 = 1 ε 123 = 1 epsi^(123)=1\varepsilon^{123}=1ε123=1. Consider the p = 2 p = 2 p=2p=2p=2-form A ~ = d x 2 d x 3 A ~ = d x 2 d x 3 tilde(A)=dx^(2)^^dx^(3)\tilde{\boldsymbol{A}}=\boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3}A~=dx2dx3, which has components
A μ ν = A ~ ( e μ , e ν ) = d x 2 ( e μ ) d x 3 ( e ν ) d x 3 ( e μ ) d x 2 ( e ν ) A μ ν = A ~ e μ , e ν = d x 2 e μ d x 3 e ν d x 3 e μ d x 2 e ν A_(mu nu)= tilde(A)(e_(mu),e_(nu))=dx^(2)(e_(mu))ox dx^(3)(e_(nu))-dx^(3)(e_(mu))ox dx^(2)(e_(nu))A_{\mu \nu}=\tilde{\boldsymbol{A}}\left(\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right)=\boldsymbol{d} x^{2}\left(\boldsymbol{e}_{\mu}\right) \otimes \boldsymbol{d} x^{3}\left(\boldsymbol{e}_{\nu}\right)-\boldsymbol{d} x^{3}\left(\boldsymbol{e}_{\mu}\right) \otimes \boldsymbol{d} x^{2}\left(\boldsymbol{e}_{\nu}\right)Aμν=A~(eμ,eν)=dx2(eμ)dx3(eν)dx3(eμ)dx2(eν)
(37.28) = δ μ 2 δ ν 3 δ μ 3 δ ν 2 (37.28) = δ μ 2 δ ν 3 δ μ 3 δ ν 2 {:(37.28)=delta_(mu)^(2)delta_(nu)^(3)-delta_(mu)^(3)delta_(nu)^(2):}\begin{equation*} =\delta_{\mu}^{2} \delta_{\nu}^{3}-\delta_{\mu}^{3} \delta_{\nu}^{2} \tag{37.28} \end{equation*}(37.28)=δμ2δν3δμ3δν2
This ( p = 2 ) ( p = 2 ) (p=2)(p=2)(p=2)-form is dual to an object with valence ( q , 0 ) = ( 3 2 , 0 ) = ( 1 , 0 ) ( q , 0 ) = ( 3 2 , 0 ) = ( 1 , 0 ) (q,0)=(3-2,0)=(1,0)(q, 0)=(3-2,0)=(1,0)(q,0)=(32,0)=(1,0), that is, a vector. The components of the vector S = A ~ S = A ~ S=*** tilde(A)\boldsymbol{S}=\star \tilde{\boldsymbol{A}}S=A~ are
(37.29) S α = 1 p ! ω μ ν i A μ ν = 1 2 ! ε μ ν α ( δ μ 2 δ ν 3 δ μ 3 δ ν 2 ) (37.29) S α = 1 p ! ω μ ν i A μ ν = 1 2 ! ε μ ν α δ μ 2 δ ν 3 δ μ 3 δ ν 2 {:(37.29)S^(alpha)=(1)/(p!)omega^(mu nu i)A_(mu nu)=(1)/(2!)epsi^(mu nu alpha)(delta_(mu)^(2)delta_(nu)^(3)-delta_(mu)^(3)delta_(nu)^(2)):}\begin{equation*} S^{\alpha}=\frac{1}{p!} \omega^{\mu \nu i} A_{\mu \nu}=\frac{1}{2!} \varepsilon^{\mu \nu \alpha}\left(\delta_{\mu}^{2} \delta_{\nu}^{3}-\delta_{\mu}^{3} \delta_{\nu}^{2}\right) \tag{37.29} \end{equation*}(37.29)Sα=1p!ωμνiAμν=12!εμνα(δμ2δν3δμ3δν2)
We conclude
(37.30) S 1 = 1 2 ( ε 231 ε 321 ) = 1 (37.30) S 1 = 1 2 ε 231 ε 321 = 1 {:(37.30)S^(1)=(1)/(2)(epsi^(231)-epsi^(321))=1:}\begin{equation*} S^{1}=\frac{1}{2}\left(\varepsilon^{231}-\varepsilon^{321}\right)=1 \tag{37.30} \end{equation*}(37.30)S1=12(ε231ε321)=1
with other components vanishing. We have, therefore, S = e 1 S = e 1 S=e_(1)\boldsymbol{S}=\boldsymbol{e}_{1}S=e1. In brief, ( d y d z ) = ( d y d z ) = ***(dy^^dz)=\star(\boldsymbol{d} y \wedge \boldsymbol{d} z)=(dydz)= e x e x e_(x)e_{x}ex in three dimensions.
Most of our work will be in Minkowski space, similar to the next example.

Example 37.7

In n = 4 n = 4 n=4n=4n=4-dimensional Minkowski space, we have the volume form ω ~ = d t d x d y d z ω ~ = d t d x d y d z tilde(omega)=dt^^dx^^dy^^dz\tilde{\boldsymbol{\omega}}=\boldsymbol{d} t \wedge \boldsymbol{d} x \wedge \boldsymbol{d} y \wedge \boldsymbol{d} zω~=dtdxdydz, with components ε μ ν α β ε μ ν α β epsi_(mu nu alpha beta)\varepsilon_{\mu \nu \alpha \beta}εμναβ. A p = 2 p = 2 p=2p=2p=2-form M ~ = B d y d z M ~ = B d y d z tilde(M)=Bdy^^dz\tilde{\boldsymbol{M}}=B \boldsymbol{d} y \wedge \boldsymbol{d} zM~=Bdydz has components
(37.35) M μ ν = M ~ ( e μ , e ν ) = B ( δ μ 2 δ ν 3 δ μ 3 δ ν 2 ) (37.35) M μ ν = M ~ e μ , e ν = B δ μ 2 δ ν 3 δ μ 3 δ ν 2 {:(37.35)M_(mu nu)= tilde(M)(e_(mu),e_(nu))=B(delta_(mu)^(2)delta_(nu)^(3)-delta_(mu)^(3)delta_(nu)^(2)):}\begin{equation*} M_{\mu \nu}=\tilde{\boldsymbol{M}}\left(\boldsymbol{e}_{\mu}, \boldsymbol{e}_{\nu}\right)=B\left(\delta_{\mu}^{2} \delta_{\nu}^{3}-\delta_{\mu}^{3} \delta_{\nu}^{2}\right) \tag{37.35} \end{equation*}(37.35)Mμν=M~(eμ,eν)=B(δμ2δν3δμ3δν2)
which is to say that M 23 = M 32 = B M 23 = M 32 = B M_(23)=-M_(32)=BM_{23}=-M_{32}=BM23=M32=B and all other components vanish. The dual to M M M\boldsymbol{M}M is a ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor M = M M = M ***M=M\star \boldsymbol{M}=\boldsymbol{M}M=M whose components are
M α β = 1 p ! ε μ ν α β M μ ν (37.36) = B 2 ε μ ν α β ( δ μ 2 δ ν 3 δ μ 3 δ ν 2 ) = B 2 ( ε 23 α β ε 32 α β ) = B ε 23 α β M α β = 1 p ! ε μ ν α β M μ ν (37.36) = B 2 ε μ ν α β δ μ 2 δ ν 3 δ μ 3 δ ν 2 = B 2 ε 23 α β ε 32 α β = B ε 23 α β {:[M^(alpha beta)=(1)/(p!)epsi^(mu nu alpha beta)M_(mu nu)],[(37.36)=(B)/(2)epsi^(mu nu alpha beta)(delta_(mu)^(2)delta_(nu)^(3)-delta_(mu)^(3)delta_(nu)^(2))],[=(B)/(2)(epsi^(23 alpha beta)-epsi^(32 alpha beta))=Bepsi^(23 alpha beta)]:}\begin{align*} M^{\alpha \beta} & =\frac{1}{p!} \varepsilon^{\mu \nu \alpha \beta} M_{\mu \nu} \\ & =\frac{B}{2} \varepsilon^{\mu \nu \alpha \beta}\left(\delta_{\mu}^{2} \delta_{\nu}^{3}-\delta_{\mu}^{3} \delta_{\nu}^{2}\right) \tag{37.36}\\ & =\frac{B}{2}\left(\varepsilon^{23 \alpha \beta}-\varepsilon^{32 \alpha \beta}\right)=B \varepsilon^{23 \alpha \beta} \end{align*}Mαβ=1p!εμναβMμν(37.36)=B2εμναβ(δμ2δν3δμ3δν2)=B2(ε23αβε32αβ)=Bε23αβ
We conclude that the only non-zero components are M 01 = M 10 M 01 = M 10 M^(01)=-M^(10)M^{01}=-M^{10}M01=M10. So we have the bivector
(37.37) M = B e x e t (37.37) M = B e x e t {:(37.37)M=Be_(x)^^e_(t):}\begin{equation*} \boldsymbol{M}=B \boldsymbol{e}_{x} \wedge \boldsymbol{e}_{t} \tag{37.37} \end{equation*}(37.37)M=Bexet
We can also see how ( d y d z ) = e x e t ( d y d z ) = e x e t ***(dy^^dz)=e_(x)^^e_(t)\star(\boldsymbol{d} y \wedge \boldsymbol{d} z)=\boldsymbol{e}_{x} \wedge \boldsymbol{e}_{t}(dydz)=exet.
Some useful examples of commonly encountered duals are collected in the margin.

Example 37.8

Most people are already familiar with a vector that results from taking a dual: the cross product. Consider the 2 -form u ~ v ~ u ~ v ~ tilde(u)^^ tilde(v)\tilde{\boldsymbol{u}} \wedge \tilde{\boldsymbol{v}}u~v~ in Euclidean n = 3 n = 3 n=3n=3n=3-dimensional space. It has components w i j = u i v j u j v i w i j = u i v j u j v i w_(ij)=u_(i)v_(j)-u_(j)v_(i)w_{i j}=u_{i} v_{j}-u_{j} v_{i}wij=uivjujvi. Its dual is a q = 1 q = 1 q=1q=1q=1 vector t t ttt, whose first component is
t 1 = 1 p ! ε μ ν 1 ( u μ v ν v μ u ν ) (37.38) = 1 2 ! ε 231 ( u 2 v 3 v 2 u 3 ) + 1 2 ! ε 321 ( u 3 v 2 v 3 u 2 ) = u 2 v 3 u 3 v 2 t 1 = 1 p ! ε μ ν 1 u μ v ν v μ u ν (37.38) = 1 2 ! ε 231 u 2 v 3 v 2 u 3 + 1 2 ! ε 321 u 3 v 2 v 3 u 2 = u 2 v 3 u 3 v 2 {:[t^(1)=(1)/(p!)epsi^(mu nu1)(u_(mu)v_(nu)-v_(mu)u_(nu))],[(37.38)=(1)/(2!)epsi^(231)(u_(2)v_(3)-v_(2)u_(3))+(1)/(2!)epsi^(321)(u_(3)v_(2)-v_(3)u_(2))],[=u_(2)v_(3)-u_(3)v_(2)]:}\begin{align*} t^{1} & =\frac{1}{p!} \varepsilon^{\mu \nu 1}\left(u_{\mu} v_{\nu}-v_{\mu} u_{\nu}\right) \\ & =\frac{1}{2!} \varepsilon^{231}\left(u_{2} v_{3}-v_{2} u_{3}\right)+\frac{1}{2!} \varepsilon^{321}\left(u_{3} v_{2}-v_{3} u_{2}\right) \tag{37.38}\\ & =u_{2} v_{3}-u_{3} v_{2} \end{align*}t1=1p!εμν1(uμvνvμuν)(37.38)=12!ε231(u2v3v2u3)+12!ε321(u3v2v3u2)=u2v3u3v2
We also obtain
(37.39) t 2 = u 1 v 3 v 1 u 3 , t 3 = u 1 v 2 v 1 u 2 (37.39) t 2 = u 1 v 3 v 1 u 3 , t 3 = u 1 v 2 v 1 u 2 {:[(37.39)t^(2)=u_(1)v_(3)-v_(1)u_(3)","],[t^(3)=u_(1)v_(2)-v_(1)u_(2)]:}\begin{align*} t^{2} & =u_{1} v_{3}-v_{1} u_{3}, \tag{37.39}\\ t^{3} & =u_{1} v_{2}-v_{1} u_{2} \end{align*}(37.39)t2=u1v3v1u3,t3=u1v2v1u2
Since, in ordinary Euclidean 3 -space, there is no difference between and up and down index, we have recreated the cross product vector
(37.40) t = u × v (37.40) t = u × v {:(37.40)t=u xx v:}\begin{equation*} t=u \times v \tag{37.40} \end{equation*}(37.40)t=u×v
where u u u\boldsymbol{u}u and v v v\boldsymbol{v}v are the vector versions of the corresponding 1 -forms.
Fig. 37.1 The parallelogram formed by vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v.
As discussed the exercises, we have for the case of n n nnn dimensions for a p p ppp-form B ~ B ~ tilde(B)\tilde{\boldsymbol{B}}B~ and q q qqq-vector T T T\boldsymbol{T}T that
B ~ = ( 1 ) s + p ( n p ) B ~ (37.41) T = ( 1 ) s + q ( n q ) T B ~ = ( 1 ) s + p ( n p ) B ~ (37.41) T = ( 1 ) s + q ( n q ) T {:[****** tilde(B)=(-1)^(s+p(n-p)) tilde(B)],[(37.41)******T=(-1)^(s+q(n-q))T]:}\begin{align*} & \star \star \tilde{\boldsymbol{B}}=(-1)^{s+p(n-p)} \tilde{\boldsymbol{B}} \\ & \star \star \boldsymbol{T}=(-1)^{s+q(n-q)} \boldsymbol{T} \tag{37.41} \end{align*}B~=(1)s+p(np)B~(37.41)T=(1)s+q(nq)T
That is, a dual of a dual is, up to a sign, the tensor we started with.

37.3 Volume forms

After that interlude into the abstract mathematics of the mappings between forms and vectors, we turn to the business of calculating areas and volumes in four-dimensional space. This can be done quite simply using the volume tensor discussed in the previous section.
In Chapter 32, we considered n = 2 n = 2 n=2n=2n=2-dimensional Euclidean space and the case shown in Fig. 37.1, where the area of the parallelogram is given by
(37.42) ( Area ) = | u x u y v x v y | = u x v y v x u y = ε μ ν u μ v ν (37.42) (  Area  ) = u x u y v x v y = u x v y v x u y = ε μ ν u μ v ν {:(37.42)(" Area ")=|[u^(x),u^(y)],[v^(x),v^(y)]|=u^(x)v^(y)-v^(x)u^(y)=epsi_(mu nu)u^(mu)v^(nu):}(\text { Area })=\left|\begin{array}{cc} u^{x} & u^{y} \tag{37.42}\\ v^{x} & v^{y} \end{array}\right|=u^{x} v^{y}-v^{x} u^{y}=\varepsilon_{\mu \nu} u^{\mu} v^{\nu}(37.42)( Area )=|uxuyvxvy|=uxvyvxuy=εμνuμvν
There are two interesting observations to make about this. The first is that the area results from inserting vectors u u u\boldsymbol{u}u and v v v\boldsymbol{v}v into a twodimensional volume tensor ω ~ ( ) = , d x d y ω ~ ( ) = , d x d y tilde(omega)()=,dx^^dy\tilde{\boldsymbol{\omega}}()=,\boldsymbol{d} x \wedge \boldsymbol{d} yω~()=,dxdy, which has components ω i j = ε i j ω i j = ε i j omega_(ij)=epsi_(ij)\omega_{i j}=\varepsilon_{i j}ωij=εij. That is
(37.43) ( Area ) = ω ~ ( u , v ) (37.43) (  Area  ) = ω ~ ( u , v ) {:(37.43)(" Area ")= tilde(omega)(u","v):}\begin{equation*} (\text { Area })=\tilde{\boldsymbol{\omega}}(\boldsymbol{u}, \boldsymbol{v}) \tag{37.43} \end{equation*}(37.43)( Area )=ω~(u,v)
The second is that, since the dual of a bivector is a number for n = 2 n = 2 n=2n=2n=2, we can link the area (a number) to the (2,0) bivector b = u v b = u v b=u^^v\boldsymbol{b}=\boldsymbol{u} \wedge \boldsymbol{v}b=uv
(37.44) ( Area ) = b = ( u v ) (37.44) (  Area  ) = b = ( u v ) {:(37.44)(" Area ")=***b=***(u^^v):}\begin{equation*} (\text { Area })=\star \boldsymbol{b}=\star(\boldsymbol{u} \wedge \boldsymbol{v}) \tag{37.44} \end{equation*}(37.44)( Area )=b=(uv)
We demonstrate this latter feature in the next example.

Example 37.9

We are working in (Euclidean) n = 2 n = 2 n=2n=2n=2 space and so a dual of a ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) tensor gives us a 0 -form (or number). The bivector has components b μ ν = u μ v ν v μ u ν b μ ν = u μ v ν v μ u ν b^(mu nu)=u^(mu)v^(nu)-v^(mu)u^(nu)b^{\mu \nu}=u^{\mu} v^{\nu}-v^{\mu} u^{\nu}bμν=uμvνvμuν, and so the dual A A AAA has components
(37.45) A = 1 2 ε μ ν ( u μ v ν v μ u ν ) = ε μ ν u μ v ν , (37.45) A = 1 2 ε μ ν u μ v ν v μ u ν = ε μ ν u μ v ν , {:(37.45)A=(1)/(2)*epsi_(mu nu)(u^(mu)v^(nu)-v^(mu)u^(nu))=epsi_(mu nu)u^(mu)v^(nu)",":}\begin{equation*} A=\frac{1}{2} \cdot \varepsilon_{\mu \nu}\left(u^{\mu} v^{\nu}-v^{\mu} u^{\nu}\right)=\varepsilon_{\mu \nu} u^{\mu} v^{\nu}, \tag{37.45} \end{equation*}(37.45)A=12εμν(uμvνvμuν)=εμνuμvν,
which is the expression for the area we had above.
We conclude that the volume of the parallelogram in n = 2 n = 2 n=2n=2n=2 space (also known as the area) is the dual of the bivector formed from its sides.
In the same way, a 4 -volume V V V\mathcal{V}V in (3+1)-dimensional spacetime can be determined by filling in slots in a volume 4 -form. As might be expected ω ~ ( A , B , C , D ) ω ~ ( A , B , C , D ) tilde(omega)(A,B,C,D)\tilde{\boldsymbol{\omega}}(\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}, \boldsymbol{D})ω~(A,B,C,D) outputs the 4 -volume V V V\mathcal{V}V of a four-dimensional
parallelepiped with sides formed from the 4 -vectors A , B , C A , B , C A,B,C\boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}A,B,C and D D D\boldsymbol{D}D. In terms of components in Minkowski spacetime, this is written as
(37.46) V = ε μ ν α β A μ B ν C α D β = det | A 0 A 1 A 2 A 3 B 0 B 1 B 2 B 3 C 0 C 1 C 2 C 3 D 0 D 1 D 2 D 3 | (37.46) V = ε μ ν α β A μ B ν C α D β = det A 0 A 1 A 2 A 3 B 0 B 1 B 2 B 3 C 0 C 1 C 2 C 3 D 0 D 1 D 2 D 3 {:(37.46)V=epsi_(mu nu alpha beta)A^(mu)B^(nu)C^(alpha)D^(beta)=det|[A^(0),A^(1),A^(2),A^(3)],[B^(0),B^(1),B^(2),B^(3)],[C^(0),C^(1),C^(2),C^(3)],[D^(0),D^(1),D^(2),D^(3)]|:}\mathcal{V}=\varepsilon_{\mu \nu \alpha \beta} A^{\mu} B^{\nu} C^{\alpha} D^{\beta}=\operatorname{det}\left|\begin{array}{cccc} A^{0} & A^{1} & A^{2} & A^{3} \tag{37.46}\\ B^{0} & B^{1} & B^{2} & B^{3} \\ C^{0} & C^{1} & C^{2} & C^{3} \\ D^{0} & D^{1} & D^{2} & D^{3} \end{array}\right|(37.46)V=εμναβAμBνCαDβ=det|A0A1A2A3B0B1B2B3C0C1C2C3D0D1D2D3|
Example 37.10
Recall that in a coordinate representation, the volume 4 -form for a metric space is
(37.47) ω ~ = g d x 0 d x 1 d x 2 d x 3 (37.47) ω ~ = g d x 0 d x 1 d x 2 d x 3 {:(37.47) tilde(omega)=sqrt(-g)dx^(0)^^dx^(1)^^dx^(2)^^dx^(3):}\begin{equation*} \tilde{\boldsymbol{\omega}}=\sqrt{-g} \boldsymbol{d} x^{0} \wedge \boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3} \tag{37.47} \end{equation*}(37.47)ω~=gdx0dx1dx2dx3
For an infinitesimal box with sides described by vectors d x 0 e 0 , d x 1 e 1 , d x 2 e 2 d x 0 e 0 , d x 1 e 1 , d x 2 e 2 dx^(0)e_(0),dx^(1)e_(1),dx^(2)e_(2)\mathrm{d} x^{0} \boldsymbol{e}_{0}, \mathrm{~d} x^{1} \boldsymbol{e}_{1}, \mathrm{~d} x^{2} \boldsymbol{e}_{2}dx0e0, dx1e1, dx2e2 and d x 3 e 3 d x 3 e 3 dx^(3)e_(3)\mathrm{d} x^{3} e_{3}dx3e3, we have an infinitesimal 4 -volume
(37.48) d V = g d x 0 d x 1 d x 2 d x 3 (37.48) d V = g d x 0 d x 1 d x 2 d x 3 {:(37.48)dV=sqrt(-g)dx^(0)dx^(1)dx^(2)dx^(3):}\begin{equation*} \mathrm{d} \mathcal{V}=\sqrt{-g} \mathrm{~d} x^{0} \mathrm{~d} x^{1} \mathrm{~d} x^{2} \mathrm{~d} x^{3} \tag{37.48} \end{equation*}(37.48)dV=g dx0 dx1 dx2 dx3
This means that an infinitesimal volume element (useful for integration) should be written as d V = g d 4 x d V = g d 4 x dV=sqrt(-g)d^(4)x\mathrm{d} \mathcal{V}=\sqrt{-g} \mathrm{~d}^{4} xdV=g d4x, where d 4 x d 4 x d^(4)x\mathrm{d}^{4} xd4x is the usual Cartesian flat-space volume element d x 0 d x 1 d x 2 d x 3 d x 0 d x 1 d x 2 d x 3 dx^(0)dx^(1)dx^(2)dx^(3)\mathrm{d} x^{0} \mathrm{~d} x^{1} \mathrm{~d} x^{2} \mathrm{~d} x^{3}dx0 dx1 dx2 dx3.
We also see that we have access to the same volume V V V\mathcal{V}V by taking the dual of the tetravector ( A B C D ) ( A B C D ) (A^^B^^C^^D)(\boldsymbol{A} \wedge \boldsymbol{B} \wedge \boldsymbol{C} \wedge \boldsymbol{D})(ABCD), which is to say
(37.49) V = ( A B C D ) (37.49) V = ( A B C D ) {:(37.49)V=***(A^^B^^C^^D):}\begin{equation*} \mathcal{V}=\star(A \wedge B \wedge C \wedge D) \tag{37.49} \end{equation*}(37.49)V=(ABCD)
Another useful object is the 3 -volume. 10 10 ^(10){ }^{10}10 Still working in (3+1)dimensional spacetime, this quantity can be represented using a dual of a trivector by defining a 3 -volume 1 1 1\mathbf{1}1-form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~. Noting that the dual of a (3,0) tensor A B C A B C A^^B^^C\boldsymbol{A} \wedge \boldsymbol{B} \wedge \boldsymbol{C}ABC is a 1 -form in four-dimensional spacetime, we define σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ to be given by
(37.50) σ ~ = ( A B C ) (37.50) σ ~ = ( A B C ) {:(37.50) tilde(sigma)=-***(A^^B^^C):}\begin{equation*} \tilde{\sigma}=-\star(A \wedge B \wedge C) \tag{37.50} \end{equation*}(37.50)σ~=(ABC)
Why define it like this? It's because this fits nicely with our use of the volume 4 -form. In fact, filling in three slots of the volume 4 -form gives us the 3 -volume 1 -form:
(37.51) σ ~ ( ) = ω ~ ( , A , B , C ) (37.51) σ ~ ( ) = ω ~ ( , A , B , C ) {:(37.51) tilde(sigma)()= tilde(omega)(","A","B","C):}\begin{equation*} \tilde{\boldsymbol{\sigma}}()=\tilde{\boldsymbol{\omega}}(, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}) \tag{37.51} \end{equation*}(37.51)σ~()=ω~(,A,B,C)
with components in Minkowski space of
(37.52) σ μ = ω ~ ( e μ , A , B , C ) = ε μ α β γ A α B β C γ (37.52) σ μ = ω ~ e μ , A , B , C = ε μ α β γ A α B β C γ {:(37.52)sigma_(mu)= tilde(omega)(e_(mu),A,B,C)=epsi_(mu alpha beta gamma)A^(alpha)B^(beta)C^(gamma):}\begin{equation*} \sigma_{\mu}=\tilde{\boldsymbol{\omega}}\left(\boldsymbol{e}_{\mu}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C}\right)=\varepsilon_{\mu \alpha \beta \gamma} A^{\alpha} B^{\beta} C^{\gamma} \tag{37.52} \end{equation*}(37.52)σμ=ω~(eμ,A,B,C)=εμαβγAαBβCγ
We see how this can be used in the next example.

Example 37.11

Consider Minkowski spacetime. A cuboid box has sides parallel to the spacelike basis vectors e i e i e_(i)\boldsymbol{e}_{i}ei, with lengths δ x , δ y δ x , δ y delta x,delta y\delta x, \delta yδx,δy and δ z δ z delta z\delta zδz. The (spacelike) 3 -volume of the box in its rest frame is given by
(37.53) σ 0 = ω ~ ( e 0 , δ x 1 e 1 , δ x 2 e 2 , δ x 3 e 3 ) = ε 0123 δ x 1 δ x 2 δ x 3 = δ x δ y δ z (37.53) σ 0 = ω ~ e 0 , δ x 1 e 1 , δ x 2 e 2 , δ x 3 e 3 = ε 0123 δ x 1 δ x 2 δ x 3 = δ x δ y δ z {:(37.53)sigma_(0)= tilde(omega)(e_(0),deltax^(1)e_(1),deltax^(2)e_(2),deltax^(3)e_(3))=epsi_(0123)deltax^(1)deltax^(2)deltax^(3)=delta x delta y delta z:}\begin{equation*} \sigma_{0}=\tilde{\boldsymbol{\omega}}\left(\boldsymbol{e}_{0}, \delta x^{1} \boldsymbol{e}_{1}, \delta x^{2} \boldsymbol{e}_{2}, \delta x^{3} e_{3}\right)=\varepsilon_{0123} \delta x^{1} \delta x^{2} \delta x^{3}=\delta x \delta y \delta z \tag{37.53} \end{equation*}(37.53)σ0=ω~(e0,δx1e1,δx2e2,δx3e3)=ε0123δx1δx2δx3=δxδyδz
which is the expected answer for a 3 -volume of a box. All of the other components of σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ vanish.
10 10 ^(10){ }^{10}10 The 3-volume of an object in (3+1)dimensional space can be thought of as a spacelike hypersurface. That is, a slice of four-dimensional spacetime at a constant time. The normal to a spacelike hypersurface is timelike. We can like hypersurface is timelike. We can
also identify 3 -volumes that are timealso identify 3 -volumes that are time-
like hypersurfaces, which are slices of like hypersurfaces, which are slices of
spacetime taken at a constant value of spacetime taken at a constant value of
one spatial coordinate. These are defined by their normals being spacelike.
11 11 ^(11){ }^{11}11 Recall that an observer with velocity u u u\boldsymbol{u}u measures an energy E = p ~ ( u ) E = p ~ ( u ) E=- tilde(p)(u)E=-\tilde{\boldsymbol{p}}(\boldsymbol{u})E=p~(u) for a particle with momentum p p p\boldsymbol{p}p a number density of particles n = J ~ ( u ) n = J ~ ( u ) n=- tilde(J)(u)n=-\tilde{\boldsymbol{J}}(\boldsymbol{u})n=J~(u). We can add that they measure a 3 -volume V = σ ~ ( u ) V = σ ~ ( u ) V= tilde(sigma)(u)V=\tilde{\boldsymbol{\sigma}}(\boldsymbol{u})V=σ~(u).
12 12 ^(12){ }^{12}12 One way to think of this sign is that it reflects the fact that if a surface moves in some direction and passes through dust particles, then in the rest frame of the surface, the flux of dust particles is directed towards the surface. Since we define the positive sense of flux through a surface as directed being outwards, a surface as directed being outwards this gives us a minus sign. Another
way to think about this is to characterize the three-dimensional hypersurfaces ize the three-dimensional hypersurfaces
using the direction of their normal, simusing the direction of their normal, sim-
ilar to the way we write d S = n d S d S = n d S d vec(S)= vec(n)dS\mathrm{d} \vec{S}=\vec{n} \mathrm{~d} SdS=n dS for ilar to the way we write d S = n d S d S = n d S d vec(S)= vec(n)dS\mathrm{d} \vec{S}=\vec{n} \mathrm{~d} SdS=n dS for
a two-dimensional surface. From our a two-dimensional surface. From our
definitions the signs are accounted for if we have that spacelike hypersurfaces have outward-directed (timelike) normals, while timelike hypersurfaces have inward directed ones (spacelike) normals. This becomes important when choosing the orientation of a surface over which to integrate in later chapters.
In the local rest frame of an observer, their velocity vector u u u\boldsymbol{u}u is taken to be equal to e 0 ^ e 0 ^ e_( hat(0))\boldsymbol{e}_{\hat{0}}e0^. As a result, we can say that the 3 -volume V V VVV of a parallelepiped with sides A , B A , B A,B\boldsymbol{A}, \boldsymbol{B}A,B and C C C\boldsymbol{C}C measured by an observer with velocity u u u\boldsymbol{u}u is 11 11 ^(11){ }^{11}11
(37.54) V = ω ~ ( u , A , B , C ) = σ ~ ( u ) . (37.54) V = ω ~ ( u , A , B , C ) = σ ~ ( u ) . {:(37.54)V= tilde(omega)(u","A","B","C)= tilde(sigma)(u).:}\begin{equation*} V=\tilde{\boldsymbol{\omega}}(\boldsymbol{u}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C})=\tilde{\boldsymbol{\sigma}}(\boldsymbol{u}) . \tag{37.54} \end{equation*}(37.54)V=ω~(u,A,B,C)=σ~(u).
While the 0th component of σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ gives the three-volume of a spacelike surface, the other components tell us about the 3 -volume of timelike surfaces.

Example 37.12

Again in Minkowski space, consider that spatial surface of our cuboid box with lengths δ y e 2 δ y e 2 delta ye_(2)\delta y e_{2}δye2 and δ z e 3 δ z e 3 delta ze_(3)\delta z e_{3}δze3. When moving at a velocity u u u\boldsymbol{u}u, in a proper time δ τ δ τ delta tau\delta \tauδτ the surface sweeps out a volume whose third edge is u δ τ u δ τ u delta tau\boldsymbol{u} \delta \tauuδτ. We then write a 3 -volume 1-form
(37.55) σ ~ ( ) = ω ~ ( , δ y e 2 , δ z e 3 , u δ τ ) (37.55) σ ~ ( ) = ω ~ , δ y e 2 , δ z e 3 , u δ τ {:(37.55) tilde(sigma)()= tilde(omega)(,delta ye_(2),delta ze_(3),u delta tau):}\begin{equation*} \tilde{\boldsymbol{\sigma}}()=\tilde{\boldsymbol{\omega}}\left(, \delta y \boldsymbol{e}_{2}, \delta z \boldsymbol{e}_{3}, \boldsymbol{u} \delta \tau\right) \tag{37.55} \end{equation*}(37.55)σ~()=ω~(,δye2,δze3,uδτ)
In the rest frame of the box, we then have u = e 0 u = e 0 u=e_(0)\boldsymbol{u}=\boldsymbol{e}_{0}u=e0 and so the components of σ ~ ( ) σ ~ ( ) tilde(sigma)()\tilde{\boldsymbol{\sigma}}()σ~() in this frame are given by
σ 0 = ω ~ ( e 0 , δ y e 2 , δ z e 3 , e 0 δ τ ) σ 1 = ω ~ ( e 1 , δ y e 2 , δ z e 3 , e 0 δ τ ) σ 2 = ω ~ ( e 2 , δ y e 2 , δ z e 3 , e 0 δ τ ) = 0 , σ 0 = ω ~ e 0 , δ y e 2 , δ z e 3 , e 0 δ τ σ 1 = ω ~ e 1 , δ y e 2 , δ z e 3 , e 0 δ τ σ 2 = ω ~ e 2 , δ y e 2 , δ z e 3 , e 0 δ τ = 0 , {:[sigma_(0)= tilde(omega)(e_(0),delta ye_(2),delta ze_(3),e_(0)delta tau)],[sigma_(1)= tilde(omega)(e_(1),delta ye_(2),delta ze_(3),e_(0)delta tau)],[sigma_(2)= tilde(omega)(e_(2),delta ye_(2),delta ze_(3),e_(0)delta tau)]:}=0,\begin{aligned} \sigma_{0} & =\tilde{\boldsymbol{\omega}}\left(e_{0}, \delta y \boldsymbol{e}_{2}, \delta z e_{3}, e_{0} \delta \tau\right) \\ \sigma_{1} & =\tilde{\boldsymbol{\omega}}\left(\boldsymbol{e}_{1}, \delta y \boldsymbol{e}_{2}, \delta z \boldsymbol{e}_{3}, \boldsymbol{e}_{0} \delta \tau\right) \\ \sigma_{2} & =\tilde{\boldsymbol{\omega}}\left(\boldsymbol{e}_{2}, \delta y \boldsymbol{e}_{2}, \delta z e_{3}, \boldsymbol{e}_{0} \delta \tau\right) \end{aligned}=0,σ0=ω~(e0,δye2,δze3,e0δτ)σ1=ω~(e1,δye2,δze3,e0δτ)σ2=ω~(e2,δye2,δze3,e0δτ)=0,
(37.56) σ 3 = ω ~ ( e 3 , δ y e 2 , δ z e 3 , e 0 δ τ ) = 0 (37.56) σ 3 = ω ~ e 3 , δ y e 2 , δ z e 3 , e 0 δ τ = 0 {:(37.56)sigma_(3)= tilde(omega)(e_(3),delta ye_(2),delta ze_(3),e_(0)delta tau)=0:}\begin{equation*} \sigma_{3}=\tilde{\omega}\left(e_{3}, \delta y e_{2}, \delta z e_{3}, e_{0} \delta \tau\right)=0 \tag{37.56} \end{equation*}(37.56)σ3=ω~(e3,δye2,δze3,e0δτ)=0
We conclude that σ 1 = δ y δ z δ τ σ 1 = δ y δ z δ τ sigma_(1)=-delta y delta z delta tau\sigma_{1}=-\delta y \delta z \delta \tauσ1=δyδzδτ, and the other components vanish. The minus sign might be slightly unexpected since we have chosen the surface using the usual right-handed conventions, but it follows from the fact that the surface is at rest in the frame in which we've worked out the components. 12 12 ^(12){ }^{12}12
The volume form is often used to compute volume elements as examined in the next example.
Example 37.13
Working in a coordinate frame we have
(37.57) ω ~ = g d x 0 d x 1 d x 2 d x 3 (37.57) ω ~ = g d x 0 d x 1 d x 2 d x 3 {:(37.57) tilde(omega)=sqrt(-g)dx^(0)^^dx^(1)^^dx^(2)^^dx^(3):}\begin{equation*} \tilde{\boldsymbol{\omega}}=\sqrt{-g} \boldsymbol{d} x^{0} \wedge \boldsymbol{d} x^{1} \wedge \boldsymbol{d} x^{2} \wedge \boldsymbol{d} x^{3} \tag{37.57} \end{equation*}(37.57)ω~=gdx0dx1dx2dx3
For an infinitesimal box with sides described in its rest frame by spacelike vectors d x 1 e 1 , d x 2 e 2 d x 1 e 1 , d x 2 e 2 dx^(1)e_(1),dx^(2)e_(2)\mathrm{d} x^{1} \boldsymbol{e}_{1}, \mathrm{~d} x^{2} \boldsymbol{e}_{2}dx1e1, dx2e2 and d x 3 e 3 d x 3 e 3 dx^(3)e_(3)\mathrm{d} x^{3} \boldsymbol{e}_{3}dx3e3 we have a 3 -volume 1-form integration element
(37.58) d σ ~ ( ) = g ε 0123 d x 0 ( ) d x 1 d x 2 d x 3 . (37.58) d σ ~ ( ) = g ε 0123 d x 0 ( ) d x 1 d x 2 d x 3 . {:(37.58)d tilde(sigma)()=sqrt(-g)epsi_(0123)dx^(0)()dx^(1)dx^(2)dx^(3).:}\begin{equation*} \mathrm{d} \tilde{\boldsymbol{\sigma}}()=\sqrt{-g} \varepsilon_{0123} \boldsymbol{d} x^{0}() \mathrm{d} x^{1} \mathrm{~d} x^{2} \mathrm{~d} x^{3} . \tag{37.58} \end{equation*}(37.58)dσ~()=gε0123dx0()dx1 dx2 dx3.
To turn this into a volume, insert the velocity and we find (noting ε 0123 = 1 ε 0123 = 1 epsi_(0123)=1\varepsilon_{0123}=1ε0123=1 )
d V = d σ ~ ( u ) = g d x 0 ( u ) d x 1 d x 2 d x 3 (37.59) = g u 0 d x 1 d x 2 d x 3 , d V = d σ ~ ( u ) = g d x 0 ( u ) d x 1 d x 2 d x 3 (37.59) = g u 0 d x 1 d x 2 d x 3 , {:[dV=d tilde(sigma)(u)=sqrt(-g)dx^(0)(u)dx^(1)dx^(2)dx^(3)],[(37.59)=sqrt(-g)u^(0)dx^(1)dx^(2)dx^(3)","]:}\begin{align*} \mathrm{d} V=\mathrm{d} \tilde{\boldsymbol{\sigma}}(\boldsymbol{u}) & =\sqrt{-g} \boldsymbol{d} x^{0}(\boldsymbol{u}) \mathrm{d} x^{1} \mathrm{~d} x^{2} \mathrm{~d} x^{3} \\ & =\sqrt{-g} u^{0} \mathrm{~d} x^{1} \mathrm{~d} x^{2} \mathrm{~d} x^{3}, \tag{37.59} \end{align*}dV=dσ~(u)=gdx0(u)dx1 dx2 dx3(37.59)=gu0 dx1 dx2 dx3,
13 13 ^(13){ }^{13}13 Recall for an observer at rest in a leading us to conclude that the invariant proper 3 -volume element is 13 d V = 13 d V = ^(13)dV={ }^{13} \mathrm{~d} V=13 dV= spacetime with a diagonal metric we have u 0 = ( g t t ) 1 2 u 0 = g t t 1 2 u^(0)=(-g_(tt))^(-(1)/(2))u^{0}=\left(-g_{t t}\right)^{-\frac{1}{2}}u0=(gtt)12, showing how the timelike component of the metric is effectively divided out of the determinant.
g u 0 d x 1 d x 2 d x 3 g u 0 d x 1 d x 2 d x 3 sqrt(-g)u^(0)dx^(1)dx^(2)dx^(3)\sqrt{-g} u^{0} \mathrm{~d} x^{1} \mathrm{~d} x^{2} \mathrm{~d} x^{3}gu0 dx1 dx2 dx3.
The components d σ μ d σ μ dsigma_(mu)\mathrm{d} \sigma_{\mu}dσμ of the infinitesimal 3-volume 1 -form will always be contracted against the components of a vector in expressions like J μ d σ μ = d σ ~ ( J ) J μ d σ μ = d σ ~ ( J ) J^(mu)dsigma_(mu)=d tilde(sigma)(J)J^{\mu} \mathrm{d} \sigma_{\mu}=\mathrm{d} \tilde{\boldsymbol{\sigma}}(\boldsymbol{J})Jμdσμ=dσ~(J).
Example 37.14
If the integration surface is the infinitesimal box in 3-space from above, we have a contribution
(37.60) J 0 d σ 0 = g J 0 ε 0123 d x 1 d x 2 d x 3 = g J 0 d x d y d z , (37.60) J 0 d σ 0 = g J 0 ε 0123 d x 1 d x 2 d x 3 = g J 0 d x d y d z , {:(37.60)J^(0)dsigma_(0)=sqrt(-g)J^(0)epsi_(0123)dx^(1)dx^(2)dx^(3)=sqrt(-g)J^(0)dxdydz",":}\begin{equation*} J^{0} \mathrm{~d} \sigma_{0}=\sqrt{-g} J^{0} \varepsilon_{0123} \mathrm{~d} x^{1} \mathrm{~d} x^{2} \mathrm{~d} x^{3}=\sqrt{-g} J^{0} \mathrm{~d} x \mathrm{~d} y \mathrm{~d} z, \tag{37.60} \end{equation*}(37.60)J0 dσ0=gJ0ε0123 dx1 dx2 dx3=gJ0 dx dy dz,
which computes the amount of the timelike (i.e. charge-like) component of vector J J J\boldsymbol{J}J in the box. A different choice of the three-dimensional integration surface might lead to a non-zero contribution like
(37.61) J 1 d σ 1 = g J 1 ε 1230 d x 2 d x 3 d x 0 = g J 1 d y d z d t . (37.61) J 1 d σ 1 = g J 1 ε 1230 d x 2 d x 3 d x 0 = g J 1 d y d z d t . {:(37.61)J^(1)dsigma_(1)=sqrt(-g)J^(1)epsi_(1230)dx^(2)dx^(3)dx^(0)=-sqrt(-g)J^(1)dydzdt.:}\begin{equation*} J^{1} \mathrm{~d} \sigma_{1}=\sqrt{-g} J^{1} \varepsilon_{1230} \mathrm{~d} x^{2} \mathrm{~d} x^{3} \mathrm{~d} x^{0}=-\sqrt{-g} J^{1} \mathrm{~d} y \mathrm{~d} z \mathrm{~d} t . \tag{37.61} \end{equation*}(37.61)J1 dσ1=gJ1ε1230 dx2 dx3 dx0=gJ1 dy dz dt.
which computes the flux of the J 1 J 1 J^(1)J^{1}J1 component of the current though the face of a surface with sides parallel to e 2 e 2 e_(2)\boldsymbol{e}_{2}e2 and e 3 e 3 e_(3)\boldsymbol{e}_{3}e3 in coordinate time d t d t dt\mathrm{d} tdt.
The material in this chapter will be used most extensively in Chapters 42 and 43 . In the next chapter, we continue to examine the mathematics of forms and turn to the question of how they give rise to a new insight into what an integral is.

Chapter summary

  • The Hodge star operation allows us to map between forms and antisymmetric tensors built from vectors.
  • Duality allows us to compute volumes using the volume tensor ω ~ ω ~ tilde(omega)\tilde{\boldsymbol{\omega}}ω~. This is a 4 -form in our usual ( 3 + 1 3 + 1 3+13+13+1 )-dimensional spacetime.
  • The 3 -volume of a box with sides described by vectors A , B A , B A,B\boldsymbol{A}, \boldsymbol{B}A,B and C C C\boldsymbol{C}C, measured by an observer with velocity u u u\boldsymbol{u}u, is given by σ ~ ( u ) = σ ~ ( u ) = tilde(sigma)(u)=\tilde{\boldsymbol{\sigma}}(\boldsymbol{u})=σ~(u)= ω ~ ( u , A , B , C ) ω ~ ( u , A , B , C ) tilde(omega)(u,A,B,C)\tilde{\boldsymbol{\omega}}(\boldsymbol{u}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C})ω~(u,A,B,C).

Exercises

(37.1) Consider the volume 1 -form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ describing the 3 volume of a box. Insert this into the ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) energymomentum tensor T T T\boldsymbol{T}T and we have that momentum p p p\boldsymbol{p}p is given via
(37.62) T ( , σ ~ ) = ( momentum crossing from negative to positive side of box ) (37.62) T ( , σ ~ ) = (  momentum crossing from   negative to positive side of box  ) {:(37.62)T("," tilde(sigma))=((" momentum crossing from ")/(" negative to positive side of box ")):}\begin{equation*} \boldsymbol{T}(, \tilde{\boldsymbol{\sigma}})=\binom{\text { momentum crossing from }}{\text { negative to positive side of box }} \tag{37.62} \end{equation*}(37.62)T(,σ~)=( momentum crossing from  negative to positive side of box )
An observer with velocity u u u\boldsymbol{u}u carrying the box measures its 3 -volume as V V VVV.
(a) Show that the volume 1 -form σ ~ σ ~ tilde(sigma)\tilde{\boldsymbol{\sigma}}σ~ is related to the velocity 1 -form by σ ~ = V u ~ σ ~ = V u ~ tilde(sigma)=-V tilde(u)\tilde{\boldsymbol{\sigma}}=-V \tilde{\boldsymbol{u}}σ~=Vu~.
Use this result to find expressions for: (b) the momentum and (c) the total energy, in terms of V V VVV and the components of T T T\boldsymbol{T}T and u ~ u ~ tilde(u)\tilde{\boldsymbol{u}}u~.
(37.2) The observer with velocity vector u u u\boldsymbol{u}u carries a box with sides A , B A , B A,B\boldsymbol{A}, \boldsymbol{B}A,B and C C C\boldsymbol{C}C which has permeable walls, allowing a particle current J J J\boldsymbol{J}J to enter it.
(a) Explain why the observer can resolve the current into the form
(37.63) J = n u + a A + b B + γ C (37.63) J = n u + a A + b B + γ C {:(37.63)J=nu+aA+bB+gamma C:}\begin{equation*} \boldsymbol{J}=n \boldsymbol{u}+a \boldsymbol{A}+b \boldsymbol{B}+\gamma \boldsymbol{C} \tag{37.63} \end{equation*}(37.63)J=nu+aA+bB+γC
where n n nnn is the number density of particles measured by the observer.
(b) Show that the number of particles N N NNN inside the box is
N = ω ~ ( J , A , B , C ) N = ω ~ ( J , A , B , C ) N= tilde(omega)(J,A,B,C)N=\tilde{\boldsymbol{\omega}}(\boldsymbol{J}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C})N=ω~(J,A,B,C)
(37.64)
(c) Now consider a second observer with velocity v v v\boldsymbol{v}v carrying a box with sides A , B A , B A^('),B^(')\boldsymbol{A}^{\prime}, \boldsymbol{B}^{\prime}A,B and C C C^(')\boldsymbol{C}^{\prime}C. What condition is necessary to ensure the boxes capture the same number of particles?
(37.3) (a) In a three-dimensional coordinate system with a metric, show that the first component of the curl of a vector v v v\boldsymbol{v}v can be written as
(37.65) 1 g : ( V 3 x 2 V 2 x 3 ) (37.65) 1 g : V 3 x 2 V 2 x 3 {:(37.65)(1)/(sqrtg):((delV_(3))/(delx^(2))-(delV_(2))/(delx^(3))):}\begin{equation*} \frac{1}{\sqrt{g}}:\left(\frac{\partial V_{3}}{\partial x^{2}}-\frac{\partial V_{2}}{\partial x^{3}}\right) \tag{37.65} \end{equation*}(37.65)1g:(V3x2V2x3)
(b) Show further that the curl is given, in an orthonormal frame in three dimensions by
( × v ) 1 ^ = ( g 11 g ) 1 2 [ x 3 ( g 22 1 2 v 2 ^ ) x 2 ( g 33 1 2 v 3 ^ ) ] ( × v ) 1 ^ = g 11 g 1 2 x 3 g 22 1 2 v 2 ^ x 2 g 33 1 2 v 3 ^ {:[(grad xx v)^( hat(1))=],[quad((g_(11))/(g))^((1)/(2))[(del)/(delx^(3))*(g_(22)^((1)/(2))v^( hat(2)))-(del)/(delx^(2))(g_(33)^((1)/(2))v^( hat(3)))]]:}\begin{aligned} & (\boldsymbol{\nabla} \times \boldsymbol{v})^{\hat{1}}= \\ & \quad\left(\frac{g_{11}}{g}\right)^{\frac{1}{2}}\left[\frac{\partial}{\partial x^{3}} \cdot\left(g_{22}^{\frac{1}{2}} v^{\hat{2}}\right)-\frac{\partial}{\partial x^{2}}\left(g_{33}^{\frac{1}{2}} v^{\hat{3}}\right)\right] \end{aligned}(×v)1^=(g11g)12[x3(g2212v2^)x2(g3312v3^)]
and cyclic permutations.
(37.4) Show that
(37.67) B ~ = ( 1 ) s + p ( n p ) B ~ T = ( 1 ) s + q ( n q ) T . (37.67) B ~ = ( 1 ) s + p ( n p ) B ~ T = ( 1 ) s + q ( n q ) T . {:[(37.67)****** tilde(B)=(-1)^(s+p(n-p)) tilde(B)],[******T=(-1)^(s+q(n-q))T.]:}\begin{align*} & \star \star \tilde{\boldsymbol{B}}=(-1)^{s+p(n-p)} \tilde{\boldsymbol{B}} \tag{37.67}\\ & \star \star \boldsymbol{T}=(-1)^{s+q(n-q)} \boldsymbol{T} . \end{align*}(37.67)B~=(1)s+p(np)B~T=(1)s+q(nq)T.
Hint: This proof, given in the 1980 book by Schutz, relies on manipulating the indices of the LeviCivita symbol. If in doubt, prove the first identity for n = 4 n = 4 n=4n=4n=4 and p = 2 p = 2 p=2p=2p=2 or similar, and then generalize. Once the first identity is proven, the second identity follows using very similar arguments.
(37.5) The Lie derivative gives access to a general method for computing the divergence θ v θ v theta_(v)\theta_{v}θv of a field v v v\boldsymbol{v}v. The divergence arises when we take the Lie derivative of the volume tensor ω ~ = ε | α β γ δ | d x α d x β d x γ ω ~ = ε | α β γ δ | d x α d x β d x γ tilde(omega)=epsi_(|alpha beta gamma delta|)dx^(alpha)^^dx^(beta)^^dx^(gamma)^^\tilde{\boldsymbol{\omega}}=\varepsilon_{|\alpha \beta \gamma \delta|} \boldsymbol{d} x^{\alpha} \wedge \boldsymbol{d} x^{\beta} \wedge \boldsymbol{d} x^{\gamma} \wedgeω~=ε|αβγδ|dxαdxβdxγ d x δ d x δ dx^(delta)\boldsymbol{d} x^{\delta}dxδ. We define the divergence via
(37.68) £ v ω ~ = θ v ω ~ . (37.68) £ v ω ~ = θ v ω ~ . {:(37.68)£_(v) tilde(omega)=-theta_(v) tilde(omega).:}\begin{equation*} £_{v} \tilde{\omega}=-\theta_{v} \tilde{\omega} . \tag{37.68} \end{equation*}(37.68)£vω~=θvω~.
(a) Using d ω ~ = 0 d ω ~ = 0 d tilde(omega)=0\boldsymbol{d} \tilde{\boldsymbol{\omega}}=0dω~=0 show that
(37.69) μ ω ~ = 0 (37.69) μ ω ~ = 0 {:(37.69)grad_(mu) tilde(omega)=0:}\begin{equation*} \nabla_{\mu} \tilde{\boldsymbol{\omega}}=0 \tag{37.69} \end{equation*}(37.69)μω~=0
(b) Using the previous result, show further that
( £ v ω ~ ) μ ν α β = ε σ ν α β v ; μ σ + ε μ σ α β v ; ν σ + ε μ ν σ β v ; α σ + ε μ ν α σ v ; β σ £ v ω ~ μ ν α β = ε σ ν α β v ; μ σ + ε μ σ α β v ; ν σ + ε μ ν σ β v ; α σ + ε μ ν α σ v ; β σ {:[-(£_(v)( tilde(omega)))_(mu nu alpha beta)=epsi_(sigma nu alpha beta)v_(;mu)^(sigma)+epsi_(mu sigma alpha beta)v_(;nu)^(sigma)],[+epsi_(mu nu sigma beta)v_(;alpha)^(sigma)+epsi_(mu nu alpha sigma)v_(;beta)^(sigma)]:}\begin{aligned} -\left(£_{v} \tilde{\boldsymbol{\omega}}\right)_{\mu \nu \alpha \beta}= & \varepsilon_{\sigma \nu \alpha \beta} v_{; \mu}^{\sigma}+\varepsilon_{\mu \sigma \alpha \beta} v_{; \nu}^{\sigma} \\ & +\varepsilon_{\mu \nu \sigma \beta} v_{; \alpha}^{\sigma}+\varepsilon_{\mu \nu \alpha \sigma} v_{; \beta}^{\sigma} \end{aligned}(£vω~)μναβ=εσναβv;μσ+εμσαβv;νσ+εμνσβv;ασ+εμνασv;βσ
(c) As a result, prove that the divergence is calculated using the connection through the familiar rule
(37.71) θ v = v = v ; μ μ (37.71) θ v = v = v ; μ μ {:(37.71)theta_(v)=-grad*v=-v_(;mu)^(mu):}\begin{equation*} \theta_{\boldsymbol{v}}=-\boldsymbol{\nabla} \cdot \boldsymbol{v}=-v_{; \mu}^{\mu} \tag{37.71} \end{equation*}(37.71)θv=v=v;μμ
(37.6) Consider a uniform distribution of dust. We have a congruence of curves formed from the world lines of dust particles with a corresponding current (tangent) field J J J\boldsymbol{J}J. Consider again an observer with velocity u u u\boldsymbol{u}u carrying a permeable box of volume V V VVV spanned by vectors A , B A , B A,B\boldsymbol{A}, \boldsymbol{B}A,B and C C C\boldsymbol{C}C.
(a) Explain why
(37.72) £ J J = £ J A = £ J B = £ J C = 0 . (37.72) £ J J = £ J A = £ J B = £ J C = 0 . {:(37.72)£_(J)J=£_(J)A=£_(J)B=£_(J)C=0.:}\begin{equation*} £_{J} \boldsymbol{J}=£_{J} \boldsymbol{A}=£_{J} \boldsymbol{B}=£_{J} \boldsymbol{C}=0 . \tag{37.72} \end{equation*}(37.72)£JJ=£JA=£JB=£JC=0.
(b) Show that for χ = ω ~ ( J , A , B , C ) χ = ω ~ ( J , A , B , C ) chi= tilde(omega)(J,A,B,C)\chi=\tilde{\boldsymbol{\omega}}(\boldsymbol{J}, \boldsymbol{A}, \boldsymbol{B}, \boldsymbol{C})χ=ω~(J,A,B,C), where ω ~ ω ~ tilde(omega)\tilde{\boldsymbol{\omega}}ω~ is the volume 4 -form, we have
(37.73) £ J χ = θ J χ (37.73) £ J χ = θ J χ {:(37.73)£_(J)chi=-theta_(J)chi:}\begin{equation*} £_{J} \chi=-\theta_{J} \chi \tag{37.73} \end{equation*}(37.73)£Jχ=θJχ
(c) Why is the case that, if the number of particles in the box is constant (as we would expect for a uniform distribution), then
(37.74) £ J χ = 0 ? (37.74) £ J χ = 0 ? {:(37.74)£_(J chi)=0?:}\begin{equation*} £_{J \chi}=0 ? \tag{37.74} \end{equation*}(37.74)£Jχ=0?
(d) Show finally that
(37.75) J = 0 (37.75) J = 0 {:(37.75)grad*J=0:}\begin{equation*} \nabla \cdot \boldsymbol{J}=0 \tag{37.75} \end{equation*}(37.75)J=0
See the book by Ludvigsen for a discussion of the material in this problem.